Current Electricity Test

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Current Electricity Test - 47

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Question 1
1 / -0

The length of wire $$AO$$, at null point, will be

A
$$37.5 cm$$

B
$$3.75 cm$$

C
$$75 cm$$

D
$$15 cm$$
Question 2
1 / -0

What is the current I for the circuit shown in the figure?

A
$$3A$$

B
$$1A$$

C
$$-5A$$

D
$$5A$$

Solution

total current leaving the circuit = total current entering the circuit

$$I + 4 = -3 + 2 $$

$$I= -5 A $$
Question 3
1 / -0

n identical cells, each of emf $$\varepsilon $$ and internal resistance $$r$$, are joined in series to form a closed circuit. One cell $$A$$ is joined with reversed polarity. The potential difference across each cell, except $$A$$, is

A
$$\displaystyle \frac{2\varepsilon }{n}$$

B
$$\displaystyle \frac{n -1 }{n}\varepsilon $$

C
$$\displaystyle \frac{n -2 }{n}\varepsilon $$

D
$$\displaystyle \frac{2n }{n -2}\varepsilon $$

Solution

Given,
Number of cells =n
EMF of each cell =ε
and Internal resistance of each cell =r
One cell is joined in reverse.
∴ Total EMF =n⋅ε−2⋅ε

$$i=\displaystyle \frac{(n -2)\varepsilon }{nr} $$

$$\displaystyle V_B-V_A=-ir+\varepsilon=\varepsilon-\frac{(n -2)\varepsilon }{nr}r=\varepsilon \left [ 1- \frac{n-2}{n}\right ]=\frac{2\varepsilon }{n}$$
Question 4
1 / -0

Two cells of emf $$E_1$$ and $$E_2 (E_1 > E_2)$$ are connected shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is $$300\ cm.$$ On connecting the same potentiometer between A and C, the balancing length is $$100\ cm.$$ The ratio $$E_1/ E_2$$ is

A
$$3 : 1$$

B
$$1 : 3$$

C
$$2 : 3$$

D
$$3 : 2$$

Solution

$$E_1 \propto 300, E_1 - E_2 \propto 100$$

$$\displaystyle \dfrac{E_1}{E_1 - E_2} = 3$$ or $$E_1 = 3E_1 - 3 E_2$$

or $$3 E_2 = 2 E_1$$ or $$\displaystyle \dfrac{E_1}{E_2} = \dfrac{3}{2}$$
Question 5
1 / -0

What is the p.d. across the terminals $$(V_T)$$ of a cell with emf E for open circuit ?

A
$$V_T< E$$

B
$$V_T>E$$

C
$$V_T=0$$

D
$$V_T=E$$

Solution

When the circuit is closed, the resulting current not only flows through the external circuit, but through the source (battery, generator, transformer, etc.) itself. All sources have an internal resistance, which causes an internal voltage drop, slightly reducing the voltage across the terminals. The larger the current, the larger the internal voltage drop, and the lower the terminal voltage.
When the circuit is open, no current flows. So there is no internal voltage drop, and the full voltage appears across the source's terminals.
This is why the potential difference across the terminals of a cell when connected to a circuit is slightly lesser than the emf of the cell.
Question 6
1 / -0

In an experiment of verification of Ohm's law, following observations are obtained:
Potential difference V (in volt) 0.5 1.0 1.5 2.0 2.5
Current I (in ampere) 0.2 0.4 0.6 0.8 1.0
What will be the potential difference V when the current I is 0.5 A ?

A
$$1.25 V.$$

B
$$0.5 V$$

C
$$2 V$$

D
$$1 V$$

Solution

A currentvoltage characteristic or IV curve (currentvoltage curve) is a relationship, typically represented as a chart or graph, between the electric current through a circuit, device, or material, and the corresponding voltage, or potential difference across it.
In the graph, the voltage is plotted along the y-axis and the current is plotted along the x-axis. Hence, the value of potential difference for the current of 0.5 amps along the x-axis can be directly found out from the y-axis. The line intersecting the point along the y-axis gives the corresponding potential difference.
Therefore, for 0.5 amps the potential difference is 1.25 V.
Question 7
1 / -0

A cell of e.m.f. $$\varepsilon$$ and internal resistance r is used to send current to an external resistance R. The total resistance of circuit,

A
$$\dfrac{Rr}{(R+r)}$$

B
$$\dfrac{R+r}{2}$$

C
$$Rr$$

D
$$R+r$$

Solution

Let us consider the circuit as given in the diagram below.
Batteries and cells have an internal resistance (r) which is measures in ohms. When electricity flows round a circuit the internal resistance of the cell itself resists the flow of current and so thermal (heat) energy is wasted in the cell itself.
$$\varepsilon =I(R+r)$$
where,
$$\varepsilon $$ = electromotive force in volts, V
I = current in amperes, A
R = resistance of the load in the circuit in ohms,
r = internal resistance of the cell in ohms.
The above equation can be written as $$\varepsilon =IR+Ir$$.
Hence, the total resistance in the circuit is given as R+r.
Question 8
1 / -0

Directions For Questions

A potentiometer is an ideal voltmeter since a voltmeter draws some current through thsadition, a constant current flows throughout the wire of a potentiometer using standard cell of emf $$e_1$$. The wire of potentiometer is made of uniform material and cross-sectional area, and it has uniform resistance per unit length. The potential gradient depends upon the current in the wire.
A potentiometer with a cell of emf 2 V and internal resistance 0.4 $$\omega$$ is used across the wire AB. A standard cadmium cell of emf 1.02 V gives a balance point at 66 cm length of wire. The standard cell is then replaced by a cell of unknown emf e (internal resistance r), and the balance point found similarly turns out to be 88 cm length of the wire. The length of potentiometer wire AB is 1 m.

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The value of e is

A
1.36 V

B
2.63 V

C
1.83 V

D
none

Solution

$$\displaystyle \frac{e}{1.02} = \frac{88}{66} $$ or $$e = 1.36 V$$
4

V is greater than applied emf 2 V, hence no balance point is obtained.

On connecting the resistance across e, current will flow in e due to

which terminal potential difference will be less than emf and the

balancing length will decrease.
Question 9
1 / -0

Directions For Questions

AB is a potentiometer wire of length 100 cm. The emf of cell $$E_{1}=2 V. $$ When a cell $$E_2$$ is connected across AC, where $$AC = 75 cm$$, no current flows from $$E_2$$. The internal resistance of the cell $$E_1$$ is negligible.

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Find the potential gradient along AB

A
0.1 Vcm$$^{-1}$$

B
0.03 Vcm$$^{-1}$$

C
0.04 Vcm$$^{-1}$$

D
0.02 Vcm$$^{-1}$$

Solution

$$\displaystyle k = \frac{E_1}{100} = \frac{2}{100}= 0.02 Vcm^{-1}$$
Question 10
1 / -0

A $$ 1^{\circ}C$$ rise in temperature is observed in a conductor by passing a certain current. If the current is doubled, then the rise in temperature is approximately

A
$$2.5^{\circ}C$$

B
$$4^{\circ}C$$

C
$$2^{\circ}C$$

D
$$1^{\circ}C$$

Solution

We know that heat $$H=i^2Rt $$ so $$H\propto i^2$$
So when en current i is double, heat will four times increase.
As heat is proportional to change in temperature i.e, $$H\propto \Delta T$$ so the rise in temperature will be $$4^oC$$.