Current Electricity Test - 16

In the above figure, \(6 \Omega\) and \(6 \Omega\) are connected in parallel.

The net resistance/equivalent resistance(R) of two \(6 \Omega\) resistances connected in parallel is given by:

\(\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

\(=\frac{1}{6}+\frac{1}{6}\)

\(=\frac{2}{6}\)

\(=\frac{1}{3}\)

\(R ^{\prime}=3 \Omega\)

Now \(R ^{\prime}\) and \(3 \Omega\) are in series,

\(\therefore R_{\text {net }}=R^{\prime}+3 \Omega\)

\(\Rightarrow R _{\text {net }}=3 \Omega+3 \Omega=6 \Omega\)

Hence, the correct option is (C).

We know that the direction of the electric field at every point due to an infinitely long straight uniformly charged wire must be radial (outward if λ > 0, inward if λ < 0).

Electric field lines start from positive charges and end at negative charges. If there is a single positive charge then electric field lines start from positive charge and end at infinity.

Therefore the direction of the electric field intensity at a point due to an infinitely long straight uniformly positive charged wire will be radially outward.

Hence, the correct option is (A).

The electric field produced by a point charge at a point in the electric field is given as,

\(E=\frac{k Q}{r^{2}}\)

\(\Rightarrow E \propto \frac{1}{r^{2}} \quad \ldots\)(1)

Electric potential due to a point charge at a point in the electric field is given as,

\(V=\frac{k Q}{r}\)

\(\Rightarrow V \propto \frac{1}{r}\quad \ldots\)(2)

Electric field due to an electric dipole on the equatorial line is given as,

\(E=\frac{k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(3)

Electric field due to an electric dipole on the axial line is given as,

\(E=\frac{2 k P}{r^{3}}\)

\(\Rightarrow E \propto \frac{1}{r^{3}}\quad \ldots\)(4)

From equation (1), equation (2), equation (3), and equation (4) it is clear that only the electric field produced by a point charge is inversely proportional to the square of the distance. So it follows the inverse-square law.

Hence, the correct option is (A).

Volt-ampere (VA) is a measurement of electric power in a direct current ( DC ) electrical circuit. The VA specification is also used in alternating current ( AC ) circuits, but it is less precise in this application, because it represents apparent power, which often differs from true power .

In a DC circuit, 1 VA is the equivalent of one watt (1 W). The power, P (in watts) in a DC circuit is equal to the product of the voltage V (in volts) and the current I (in amperes):

P = VI

Hence, the correct option is (C).

The longer length of the potentiometer wire is preferred for accurate measurements. The longer the wire length, the lesser is the value of fall of potential per unit length of wire resulting in a lesser value of least count (more accuracy).

Copper wire is not used in potentiometers, this is because the temperature coefficient of resistance of the copper wire is large and its resistivity is small.

In a potentiometer, Electric field intensity, \(E=\frac{V}{l}=\) constant i.e., the potential gradient is constant. Therefore, the electric field inside the potentiometer is variable.

If the length of the wire is halved, the potential gradient along the length of the potentiometer wire increases. As a result position of zero deflection will occur at a shorter length.

Hence, the correct option is (D).

Given:

\(R =\) radius of the sphere

\(E _{1}= E\)

\(q _{1}= Q\)

\(q _{2}=2\)

\(q _{1}=2 Q\)

\(r _{1}= r\)

\(r _{2}=2 r\)

Since, \(r>R\)

So the point is present outside the sphere.

We know that the electric field intensity at a point outside the metallic sphere is given as,

\(E=\frac{k q}{r^{2}} \quad \ldots\)(1)

Where, \(E=\) electric field intensity, \(k =9 \times 10^{9} N - m ^{2} / C ^{2}, q =\) charge on the sphere, and \(r =\) distance of the point from the center of the sphere

When, \(q _{1}= Q\) and \(r _{1}= r\)

\(E_{1}=\frac{k q_{1}}{r_{1}^{2}}\)

\(\Rightarrow E_{1}=\frac{k Q}{r^{2}}\quad \ldots\)(2)

When, \(q _{2}=2 Q\) and \(r _{2}=2 r\)

\(E_{2}=\frac{k q_{2}}{r_{2}^{2}}\)

\(\Rightarrow E_{2}=\frac{2 \times k Q}{(2 r)^{2}}\)

\(\Rightarrow E_{2}=\frac{k Q}{2 r^{2}}\quad \ldots\)(3)

By equation (2) and equation (3),

\(E_{2}=\frac{E_{1}}{2}\)

Since \(E_{1}=E\)

\(E_{2}=\frac{E}{2}\)

Hence, the correct option is (D).

The bridge is said to be balanced when deflection in the galvanometer is zero i.e., no current flows through the galvanometer or in other words \(V _{ B }= V _{ D }\)

In the balanced condition \(\frac{P}{Q}=\frac{R}{S}\), on mutually changing the position of cell and galvanometer, this condition will not change.

Therefore, in the circuit given, the correct relation to a balanced Wheatstone bridge is \(\frac{P}{R}=\frac{Q}{S}\).

Hence, the correct option is (C).

Nuclear force does not follow the inverse square law.

An inverse-square law is any scientific law stating that a specified physical quantity is inversely proportional to the square of the distance from the source of that physical quantity.

• The nuclear force is a force that acts between the protons and neutrons of atoms.
• The nuclear force is the force that binds the protons and neutrons in a nucleus together.
• This force can exist between protons and protons, neutrons and protons, or neutrons and neutrons.
• This force is what holds the nucleus together.
• Nuclear force does not follow the inverse square law.

Hence, the correct option is (B).

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

If, \(W=1\) Joule and \(t =1\) second, then \(P =1\) watt

Thus, the power of an electric circuit is said to be one watt or 1 ampere volt if one joule of energy is consumed in one second.

Hence, the correct option is (A).

When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.

The net resistancelequivalent resistance(R) of resistances in parallel is given by:

\(\frac{1}{R_{e f f}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

Given:

\(R_{1}=10 \Omega\) and \(R_{2}=30 \Omega\)

Since, they are parallel connected \(\frac{1}{R_{\text {eff }}}=\frac{1}{10}+\frac{1}{30}\)

\(R_{e f f}=\frac{10 \times 30}{10+30}=\frac{300}{40}=7.5 \Omega\)

Hence, the correct option is (D).

Meter Bridge: It is an electrical instruments based on the principle of Wheatstone bridge and is used to measure the resistance of a resistor.

The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stones bridge.

The formula meter bridge is given below:

ρ = Lπr2X

Where, L be the length of the wire and r be its radius.

Hence, the correct option is (C).

When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.

The net resistance/equivalent resistance(R) of resistances in parallel is given by:

\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

Suppose two resistors \(R _{1}\) and \(R _{2}\) are connected in parallel \(\left( R _{1}> R _{2}\right)\)

As we know that the voltage across both resistors is the same in parallel combination. Therefore, the current through \(R_{1}\) is:

\(I_{1}=\frac{V}{R_{1}}\)

Current through \(R _{2}\):

\(I_{2}=\frac{V}{R_{2}}\)

Since, \(R _{1}> R _{2}\),

Therefore, \(l _{1}< I _{2}\)

Hence, the correct option is (D).

Given:

\(P _{1}=200 W\)

\(P _{2}=100 W\)

\(V =100 V\)

We know that:

The rate at which the electric energy is dissipated or consumed is termed as electric power.

The electric power is given as,

\(P=I V=I^{2} R=\frac{V^{2}}{R}\)

Where, \(P =\) electric power, \(V =\) voltage, \(I =\) current and \(R =\) resistance

\(R=\frac{V^{2}}{P}\)

\(R \propto \frac{1}{P} \quad \ldots(1)\)

For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.

So, the bulb which has more power will have low resistance, therefore the \(100 W\) bulb will have more resistance compared to the \(200 W\) bulb.

The heat dissipated by the bulb is given as,

\(H = I ^{2} R\)

Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.

Since the resistance of the \(100 W\) bulb is more, so the heat dissipation of the \(100 W\) bulb will be more.

The brightness of the bulb depends on the heat dissipation by the bulb, so the \(100 W\) bulb will have more brightness.

Hence, the correct option is (A).

Given:

distance \((r)=0.2\) m

charge \((q)=1.6 \times 10^{-7}\) C

We know that:

Electric field at a point \(0.2\) m from the center of the ball,

\(E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\)

Where, \(\varepsilon_{0}=\) the permittivity of free space,

\(E=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.2)^{2}}\)

\(=\frac{14.4 \times 10^{2}}{0.04}\)

\(=360 \times 10^{2}\)

\(=3.6 \times 10^{4} N / C\)

\(\therefore\) The electric field at a point \(0.2\) m from the center of the iron ball \(=3.6 \times 10^{4} NC ^{-1}\)

Hence, the correct option is (A).

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V =\) Potential diffrence, \(R =\) Resistance and \(I =\) current

From above equation it is clear that all options are correct.

Hence, the correct option is (D).

Given:

\(I = \frac{Q}{t} =\frac{96000}{3600}=\frac{80}{3A}\)

\(V =50\) V

\(t =1\) hour \(=3600\) sec

Where, \(I =\) current, \(R =\) resistance, \(t =\) the time taken, \(V =\) electric potential and \(Q =\) quantity of charge flowing

We know that:

The amount of heat produced (H) in joules is:

\(H = Vlt\)

\(\Rightarrow H=50 \times \frac{80}{3} \times 3600\)

\(\Rightarrow H =4.8 \times 10^{6} J\)

Hence, the correct option is (C).

The process of electroplating, using the phenomenon of electrolysis, is primarily based on chemical effect of electric current.

• The phenomenon of producing heat by the electric current is called the heating effect of electric current.
• The amount of heat produced (H) in joules  = I²Rt

Where, I is current, R is resistance and t is the time taken

• Due to the heating effect of the current, the filament of the bulb gets heated to a high temperature and it starts glowing.
• Similarly, the metals are melted and coated on another material to prevent corrosion. It is a chemical effect of electric current.

Hence, the correct option is (B).

Potential gradient (k): Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e.,

\(k=\frac{V}{l}\)

Where, \(V=\) Potential difference and \(l =\) length of the wire

Potential gradient directly depends upon:

• The resistance per unit length (R/L) of potentiometer wire.
• The radius of potentiometer wire ( i.e., Area of crosssection)
• The specific resistance of the material of potentiometer wire (i.e., ρ)
• The current flowing through potentiometer wire (I).

Hence, the correct option is (A).

Here, 4 Ω across AB and AC are  in a series combination, the equivalent resistance is:

\(S = R _{1}+ R _{2}=4 \Omega+4 \Omega=8 \Omega\)

Now 4 Ω and 8 Ω are connected in parallel combination, the equivalent resistance is:

\(\frac{1}{R_{e q}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}\)

\(\Rightarrow R _{ eq }=\frac{8}{3} \Omega\)

Hence, the correct option is (D).

In a series circuit, the total resistance in the circuit is equal to the sum of all the resistances in the circuit.

• A lamp is nothing but resistance, hence with the addition of a lamp the total resistance of the circuit will increase.
• Now when the voltage is kept constant the current in the circuit will decrease with the increase of resistance in accordance with ohm’s Law (I = V/R) (with an increase in denominator value of fraction decreases).
• Thus, on adding more lamps into a series circuit, the overall current in the circuit will decrease if the voltage is kept constant. 
• The opposite will happen in a circuit where lamps/resistors will be connected parallel.

Hence, the correct option is (B).

Given:

ΔV = 2V

Where, ΔV is the change in potential.

A conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then Ohm’s law states that:

V ∝ I

V = RI

Where, R is called the resistance of the conductor. Resistance depends on the shape, size, and material of the conductor.

ΔV = Δ(RI) = RΔI

\(\Rightarrow \Delta I=\frac{\Delta(V)}{R}=\frac{2}{R}\)

We need R, the resistance to find the change in current per 2V change in the potential applied.

Hence, the correct option is (D).

The direction of the electric field at every point must be radial (outward if \(\lambda>0\), inward if \(\lambda<0\)).

We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as:

\(E=\frac{\lambda}{2 \pi \epsilon_{o} r}\)

\(\Rightarrow E \propto \lambda \quad \ldots\)(1)

Where, \(\lambda=\) linear charge density, \(\epsilon_{0}=\) permittivity, and \(r=\) distance of the point from the wire

Hence, the correct option is (B).

Given:

Potential difference \(( V )=220 V\)

Power of the bulb \(( P )=40 W\)

We know that:

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

The resistance can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{40}=1210 \Omega\)

Hence, the correct option is (A).

We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as,

\(E=\frac{\lambda}{2 \pi \epsilon_{o} r}\)

\(\Rightarrow E \propto \frac{1}{r} \quad \ldots\)(1)

Where, \(\lambda=\) linear charge density, \(\epsilon_{0}=\) permittivity, and \(r=\) distance of the point from the wire

From the above equation, it is clear that the electric field of an infinitely long straight wire is proportional to \(\frac{1}{r}\).

Hence, the correct option is (B).

1 unit of electric energy: When one-kilowatt load works for 1 hour then the energy consumed is called 1 unit of electricity.

\(1\) Unit of electricity \(=1 KWh =1000\) Watt-hour \(=3.6 \times 10^{6} J\)

1 Kilo-watt \(=1000\) Watt

1 Watt: The energy consumption rate of 1 joule per second is called 1 watt.

Since \(1 hr =3600\) sec

1 Kilo-watt \(=1000\) Watt

\(1 KWh =1000\) Watt-hour \(=1000 \times 3600\) W.s \(=3.6 \times 10^{6}\) J

Hence, the correct option is (C).

Heating effect produced by current is due to the collision of electrons.

• When a potential difference is applied across the ends of a conductor, its free electrons get accelerated in the opposite direction of the applied field.
• But the speed of the electrons does not increase beyond a constant drift speed. This is because, during the course of their motion, the electrons collide frequently with the positive metal ions.
• The kinetic energy gained by the electrons during the intervals of free acceleration between collisions is transferred to the metal ions at the time of the collision.
• The metal ions begin to vibrate about their mean positions more and more violently.

Hence, the correct option is (A).

Electric Field at a point due to a line charge distribution \(+\lambda\) at the mid distance (i.e. \(\frac{R}{2}\) ) is:

\(E_{+\lambda}=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

Similarly, Electric field due to line charge \(-\lambda\) is:

\(E_{-\lambda}=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

Both the fields have the same direction at the point A i.e. from \(+\lambda\) towards \(-\lambda\)

Thus, resultant electric field is equal to:

\(E=E_{+\lambda}+E_{-\lambda}\)

\(=\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}+\frac{\lambda}{2 \pi \varepsilon_{O}\left(\frac{R}{2}\right)}\)

\(=\frac{2 \lambda}{\pi \varepsilon_{o} R}\)

Hence, the correct option is (B).

Given:

Potential difference \(( V )=220 V _{1}\)

Power of the bulb \(( P )=100 W\)

Actual voltage \(\left( V ^{\prime}\right)=110 V\)

We know that:

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

The resistance of the bulb can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{100}=484 \Omega\)

The power consumed by the bulb.

\(P=\frac{V^{'2}}{R}=\frac{(110)^{2}}{484}=25 W\)

Hence, the correct option is (D).

The resistance of the wire is directly proportional to the length of the wire and here we are cutting the wire into n equal parts, so the new resistance will be:

\(R _{1}= R _{2}= R _{3}=\ldots \ldots \ldots \ldots R _{ n }= \frac{R}{n}\)

Now, the resistances are placed in parallel combination, therefore, the equivalent resistance will be

\(\frac{1}{R_{ eq }}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\ldots \ldots \ldots \ldots \frac{1}{R_{n}}\)

\(\frac{1}{R_{e q}}=n\left[\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \ldots \ldots \ldots \ldots +\frac{1}{R}\right]=n\left[\frac{n}{R}\right]\)

\(R_{e q}=\frac{R}{n^{2}}\)

Hence, the correct option is (D).

Given:

\(E=10^{4}\) Newton per Coulomb

\(q=1.6 \times 10^{-19}\) C

We know that:

\(F=q E\)

Where, \(F\) is the force due to the electric field, \(q\) is the charge, and \(E\) is the electric field.

\(F=\left(1.6 \times 10^{-19}\right) \times 10^{4}\)

\(\Rightarrow F=1.6 \times 10^{-15}\) Newton

Hence, the correct option is (D).