Current Electricity Questions — NEET Physics

Q1

A resistor is connected to a battery of 12 V emf and internal resistance

2 \Omega

. If the current in the circuit is 0.6 A , the terminal voltage of the battery is:

A 10 V

B 1.2 V

C 12 V

D 10.8 V

Correct Answer

Option D

Solution

Sol. Circuit can be draw as, ⇒ Terminal voltage of battery $V=E$ - ir

=12-0.6 \times 2=10.8 \mathrm{~V}

Q2

A uniform metallic wire having resistance

4 \Omega

is bent to form a square loop (ABCD) (see figure). A resistance of

2 \Omega

is connected between points

B

and

D

and a battery of 2 V is connected across points

A

and

C

as shown in the figure. Now the value of current

(l)

is :

A 2 A

B 8 A

C 4.5 A

D 4 A

Correct Answer

Option A

Solution

→ Each side will have resistance of $1 \Omega$ → It is balanced wheat stone bridge → No current in the resistance of $2 \Omega$ $\rightarrow \quad\left[R_{\text{effective }}\right]_{A C}=1 \Omega$

\begin{aligned} & \rightarrow \quad I=\frac{E}{R_{\text{eff. }}}=\frac{2}{1} \\ & \rightarrow \quad I=2 \mathrm{~A} \end{aligned}

Q3

A room heater is rated

400 \mathrm{~W}, 220 \mathrm{~V}

. If the supply voltage drops to 200 V , what will be the power consumed (approximately)?

A 200 W

B 400 W

C 331 W

D 121 W

Correct Answer

Option C

Solution

Given: Rated power $P_1 = 400\,\text{W}$ Rated voltage $V_1 = 220\,\text{V}$ New voltage $V_2 = 200\,\text{V}$ Using the relation

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

First, find the resistance of the heater:

R = \frac{V_1^2}{P_1} = \frac{220^2}{400} = \frac{48400}{400} = 121\,\Omega

Now find the new power at $200\,\text{V}$ :

P_2 = \frac{V_2^2}{R} = \frac{200^2}{121} = \frac{40000}{121} \approx 330.6\,\text{W}

So, approximately,

P_2 \approx 331\,\text{W}

Therefore, the correct answer is:

\boxed{\text{Option C: } 331\,\text{W}}

Q4

A galvanometer of resistance

100 \Omega

gives full scale deflection for a current of 1 mA . It is converted into an ammeter of range

0-10 \mathrm{~A}

. The shunt required is:

A

0.01 \Omega

B

0.10 \Omega

C

1.0 \Omega

D

0.001 \Omega

Correct Answer

Option A

Solution

\begin{aligned} i_G & =1 \mathrm{~mA}=0.001 \mathrm{~A} \\ i_S & =10-i_G \simeq 10 \mathrm{~A} \end{aligned}

Both shunt resistance and galvanometer are in parallel connection

\begin{array}{ll} \therefore & i_S r_S=i_G R_G \\ \Rightarrow & 10 \times r_s=0.001 \times 100 \\ \Rightarrow & r_s=0.01 \Omega \end{array}

Q5

In a metre bridge experiment (see figure), the positions of the cell,

E

, and galvanometer,

G

, are interchanged. We shall observe in the galvanometer:

A Only the left-sided deflection

B There will be no deflection irrespective of the position of the jockey

C Only the right-sided deflection

D Both right-sided and left-sided deflection and at balance point, no deflection

Correct Answer

Option D

Solution

Position of null point will not change when galvanometer ( $G$ ) and the cell ( $E$ ) are interchanged.

There will be no deflection in galvanometer only at balance point.

In an unbalanced meter bridge, if $E$ and $G$ are interchanged mutually, then the deflection in galvanometer may be towards left-side or right-side.

Q6

A wire of resistance

R

is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

A

\dfrac{R}{16}

B

\dfrac{R}{8}

C

\dfrac{R}{64}

D

\dfrac{R}{32}

Correct Answer

Option A

Solution

After cutting the wire into 8 equal pieces: Resistance of each piece: Each piece has a resistance $R' = \dfrac{R}{8}$ .

Parallel Combination in Each Set: Each set consists of 4 pieces connected in parallel: Resistance of each set: $R'' = \dfrac{R'}{4} = \dfrac{R}{32}$ Connecting the Sets in Series: With both sets connected in series, the equivalent resistance is calculated as: $R_{\text{eq}} = R'' + R'' = 2 \times \dfrac{R}{32} = \dfrac{R}{16}$ Thus, the net effective resistance of the combination is $\dfrac{R}{16}$ .

Q7

A constant voltage of 50 V is maintained between the points

A

and

B

of the circuit shown in the figure. The current through the branch

C D

of the circuit is:

A 2.5 A

B 3.0 A

C 1.5 A

D 2.0 A

Correct Answer

Option D

Solution

\begin{aligned} & R_{A B}=(1 \Omega / 3 \Omega) \text{ in series with }(2 \Omega / 4 \Omega) \\ & =\frac{3 \times 1}{3+1}+\frac{2 \times 4}{2+4} \\ & =\frac{3}{4}+\frac{8}{6}=\frac{9+16}{12}=\frac{25}{12} \Omega \end{aligned}

Now total current through cell

\begin{aligned} & I=\frac{50}{25 / 12}=24 \mathrm{~A} \\ & I_{1 \Omega}=\frac{3}{4} \times 24=18 \mathrm{~A}, I_{3 \Omega}=\frac{1}{4} \times 24=6 \mathrm{~A} \\ & I_{2 \Omega}=\frac{4}{6} \times 24=16 \mathrm{~A}, I_{4 \Omega}=\frac{2}{6} \times 24=8 \mathrm{~A} \end{aligned}

Using junction rule at $C, I_{C D}=18-16=2 \mathrm{~A}$ (From $C$ to $D$ )

Q8

A uniform wire of diameter

d

carries a current of

100 \mathrm{~mA}

when the mean drift velocity of electrons in the wire is

v

. For a wire of diameter

\dfrac{d}{2}

of the same material to carry a current of

200 \mathrm{~mA}

, the mean drift velocity of electrons in the wire is

A

4 v

B

8 v

C

v

D

2 v

Correct Answer

Option B

Solution

To solve this problem, we need to understand the relationship between the current, the drift velocity, and the cross-sectional area of the wire.

The electric current

I

in a wire is given by

I = n e A v_d

where:

n

is the number density of electrons,

e

is the charge of an electron,

A

is the cross-sectional area of the wire, and

v_d

is the mean drift velocity of the electrons. Let's denote the current in the thicker wire as

I_1 = 100 \mathrm{~mA} = 0.1 \mathrm{~A}

and the current in the thinner wire as

I_2 = 200 \mathrm{~mA} = 0.2 \mathrm{~A}

. The diameter of the thicker wire is

d

, so its cross-sectional area,

A_1

, is

A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

For the thinner wire, the diameter is

\frac{d}{2}

, so its cross-sectional area,

A_2

, is

A_2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16}

Using the relation for current in the thicker wire, we have

I_1 = n e A_1 v = n e \left(\frac{\pi d^2}{4}\right) v = 0.1 \mathrm{~A}

Let the mean drift velocity in the thinner wire be denoted by

v'

. For the thinner wire:

I_2 = n e A_2 v' = n e \left(\frac{\pi d^2}{16}\right) v' = 0.2 \mathrm{~A}

Now, we can equate the expressions for current and solve for

v'

:

\frac{\pi d^2}{16} n e v' = 0.2 \mathrm{~A}

v' = \frac{0.2 \mathrm{~A}}{n e \left(\frac{\pi d^2}{16}\right)}

From the equation for the thicker wire:

0.1 \mathrm{~A} = \frac{\pi d^2}{4} n e v

v = \frac{0.1 \mathrm{~A}}{n e \left(\frac{\pi d^2}{4}\right)}

Dividing the two equations, we get

\frac{v'}{v} = \frac{0.2 \mathrm{~A}}{\left(\frac{\pi d^2}{16}\right) n e} \times \frac{(\frac{\pi d^2}{4}) n e}{0.1 \mathrm{~A}} = \frac{0.2 \mathrm{~A} \times 4}{0.1 \mathrm{~A} \times 16} = \frac{4}{1} = 8

Thus,

v' = 8 v

The mean drift velocity of electrons in the thinner wire is 8 times the mean drift velocity in the thicker wire.

Therefore, the correct option is: Option B :

8 v

Q9

A uniform metal wire of length

l

has

10 \Omega

resistance. Now this wire is stretched to a length

2l

and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

A

10 \Omega

B

5 \Omega

C

40 \Omega

D

20 \Omega

Correct Answer

Option A

Solution

R_0=10 \Omega

After stretching its length upto

2l

\begin{aligned} R_1 & =n^2 R_0 \\ & =4 R_0 \\ & =40 \Omega \end{aligned}

R_{A B}=\frac{20 \times 20}{20+20}=10 \Omega

Q10

The terminal voltage of the battery, whose emf is

10 \mathrm{~V}

and internal resistance

1 \Omega

, when connected through an external resistance of

4 \Omega

as shown in the figure is:

A 4 V

B 6 V

C 8 V

D 10 V

Correct Answer

Option C

Solution

\begin{aligned} \text{ Current in circuit } i & =\frac{10}{4+1}=2 \mathrm{~A} \\ \text{ Terminal voltage } & =E-i R \\ & =10-2 \times 1=8 \mathrm{~V} \end{aligned}