Current Electricity Test

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Current Electricity Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

Express which of the following setups can be used to verify Ohm's law :-

A

B

C

D

Solution

The ammeter is always connected in series and the voltmeter is connected in parallel to the circuit. so the option B is correct.
Question 2
1 / -0

Figure $$(a)$$ below shows a Wheatstone bridge in which $$P, Q, R, S$$ are fixed resistances, $$G$$ is a galvanometer and $$B$$ is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positions of $$B$$ and $$G$$ are interchanged,. as shown in figure $$(b)$$. The new deflection of the galvanometer.

A
Is to the left.

B
Is to the right.

C
Is zero.

D
Depends on the values of $$P, Q, R, S$$.

Solution

This is one charactristic of a wheatstone bridge, that if we interchange the position of galvanometer and battery, there will no any change in the output.
Question 3
1 / -0

In the circuits shown below the ammeter $$A$$ reads $$4 amp$$ and the voltmeter $$V$$ reads $$20 volts$$. The value of the resistance $$R$$ is

A
Slightly more than 5 ohms

B
Slightly less than 5 ohms

C
Exactly 5 ohms

D
None of the above

Solution

$$i=i_{1}+i_{2}$$ (Kirchoff Juction Law)
$$i_{1}R=V$$ (ohm law)
Given $$V=20$$
$$i=4A$$
$$(i-i_{2})R=20$$
$$R=(\dfrac{20}{4-i_{2}})\Omega$$
The value of R is slightly more than $$5\Omega.$$
Question 4
1 / -0

How many electrons constitute current of one micro ampere in one second?

A
$$6.25\times10^8$$

B
$$6.25\times10^{12}$$

C
$$6.25\times10^9$$

D
$$6.25\times10^{15}$$

Solution

Given: Current ,$$ I = 1\mu A = 10^{-6}A$$
We know that $$Q = ne, t = 1s$$
Where $$n$$ is the number of electrons and e is the charge on $$1$$ electron and $$e = 1.6\times10^{-19}C$$

$$n = \displaystyle\frac{I}{e} = \displaystyle\frac{10^{-6}}{1.6\times10^{-19}} = 6.25\times10^{12} Electrons$$
Question 5
1 / -0

Figure below shows a portion of an electric circuit with the currents in ampreres and their directions. The magnitude and direction of the current in the portion PQ is :

A
0A

B
3A from P to Q

C
4A from Q to P

D
6A from Q to P

Solution

Answer is D.

The explanation is given below.
The point Q is not mentioned, However answering it by keeping the end point of P as Q.

The currents, 2 A and 8 A combine as 10 A and flows out. This 10 A is split as 6 A and flows down and then as 4 A and flows up.
The 6 A is split as 4 A and 2 A at the bottom. The 4 A is split as 3 A and 1 A in the middle.
Therefore, the 3 A and 3 A flows towards P from Q.
Hence, a magnitude of 6 A flows in the direction from Q to P.
Question 6
1 / -0

In the following circuit, each resistor has a resistance of $$15 \Omega$$ and the battery has an e.m.f. of $$12 V$$ with negligible internal resistance.
When a resistor of resistance $$R$$ is connected between $$D \& F$$, no current flows through the galvanometer (not shown in the figure) connected between $$C \& F$$. Calculate the value of $$R$$.

A
$$10 \Omega$$

B
$$15 \Omega$$

C
$$5 \Omega$$

D
$$30 \Omega$$

Solution

Voltage at point C $$\Rightarrow \frac {V_{c}-0}{15}=\frac {12-{V_{c}}}{30}$$
$$2V_c=12-V_c$$
$$3V_c=12$$
$$V_c=4volt$$
It is given that there is no any current through line CF $$\Rightarrow V_c=V_F$$
$$\Rightarrow V_F=4volt$$

$$\Rightarrow I_2= \dfrac {12-{V_{F}}}{15}=\dfrac {12-4}{15}= \dfrac {8}{15}A$$

Now ohm's law for unknown resistance $$R_1 \Rightarrow {\Delta}V=IR_1$$
$$(4-0)=(\dfrac {30}{30+R_1})(\dfrac {8}{15})R_1$$

$$4=\dfrac {16R_1}{30+R_1}$$

$$120+4R_1=16R_1$$

$$12R_1=120$$

$$R_1=10{\Omega}$$
Question 7
1 / -0

What is the minimum resistance that one can obtain by connecting all the five resistances each of $$0.5\Omega$$?

A
$$\frac{1}{10} \Omega$$

B
$$\frac{1}{5} \Omega$$

C
$$ \frac{1}{50} \Omega$$

D
$$\frac{1}{25} \Omega$$

Solution

Given $$R=0.5=\dfrac{1}{2}\Omega$$

For obtaining minimum resistance, all five resistances should be connected in parallel.
And the equivalent resistance =$$\dfrac{R}{5}$$

$$=\dfrac{0.5}{5}=\dfrac{1}{10}\Omega$$

Answer-(A)
Question 8
1 / -0

What is the effective resistance between points P and Q?

A
5 $$\Omega$$

B
10 $$\Omega$$

C
15 $$\Omega$$

D
20 $$\Omega$$

Solution

$$\textbf{Hint: Here all the resistors are connected in parallel}$$
Step1: Find the type of connection for each resistance
Here, there are 3 resistance of 12 Ω, 15 Ω and 20 Ω. From point P to Q, there are 3 paths in which all resistance are different. (i.e. current can pass from any resistance). So, all the resistance are in parallel connection with each other.
Step2: Find the net resistance of the circuit
Assume that net resistance of the circuit is R_{net}. So, from the formula of parallel resistance connection,
$$\dfrac{1}{R_{net}} = \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{20}$$

LCM of 12, 15, 20 is 60. So,
$$\dfrac{1}{R_{net}} = \dfrac{5\ +\ 4\ +\ 3}{60} = \dfrac{12}{60} = \dfrac{1}{5}$$

$$\Rightarrow R_{net} = 5\ Ω$$
$$Answer:$$
Hence, option A is the correct answer.
Question 9
1 / -0

An electric iron draws a current of 15 A from a 220 V supply, What is the cost of using iron for 30 min everyday for 15 days if the cost of unit (1 unit =1 kWhr) is 2 rupees ?

A
Rs 49.5

B
Rs 60

C
Rs 40

D
Rs 10

Solution

As we know that power(P)=v$$\times$$I
so P=15$$\times$$220=3300 watt
cost per unit=2 rupees
therefore,$$\dfrac {3300\times2\times15\times30}{60\times1000}$$=49.5 rupees.
Question 10
1 / -0

The value of equivalent resistance between the points $$A\;and\;B$$ in the given circuit, will be

A
$$6\;R$$

B
$$\displaystyle\frac{4\;R}{11}$$

C
$$\displaystyle\frac{11\;R}{4}$$

D
$$\displaystyle\frac{R}{6}$$

Solution

$$\text{Refer image-1}$$
In the given circuit resistor cd, de and ef are in series.
$$\therefore R_{eq}=cd+de+ef=R+R+R=3R$$

Now, the circuit will be like $$\text{Refer image-2}$$
$$\text{Refer image-2}$$
In this circuit cf and de are parallel
$$\therefore R_{eq}=\dfrac{R_{cf}\times R_{de}}{R_{cf}+R_{de}}$$

$$=\dfrac{R\times 3R}{R+3R}$$

$$=\dfrac{3R}{4}$$

Now, the circuit will be like $$\text{Refer image-3}$$
$$\text{Refer image-3}$$
In this circuit, $$Ac,cf,fB$$ are in series

$$\therefore R_{AB}=R+\dfrac{3R}{4}+R=2R+\dfrac{3R}{4}=\dfrac{11R}{4}$$

$$\therefore$$ Correct option is C.