Current Electricity Online Objective Test | Current Electricity Online Objective Test

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q1. Electromotive force is

A. The work done per unit charge by the source in taking the charge from lower to higher potential energy

B. The work done by the source in taking the charge from lower to higher potential energy

C. The number of charges pumped by the source from lower to higher potential energy

D. The work done per unit charge by the source in taking the charge from higher to lower potential energy

Answer : Option A
Explaination / Solution:

An electric field exists in the electrolyte between the positive and negative terminals of the battery. In the external circuit, the current flows from the positive electrode to the negative electrode. To maintain continuity, in the electrolyte, the current (positive charges) flow from the negative electrode (lower potential) to the positive electrode (higher potential). Work done by the source in taking unit positive charge from lower to higher potential is called electromotive force.

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q2. A milliammeter of range 10 mA and resistance 9 ohms is joined in a circuit as shown. The device gives full-scale deflection for current I when A and B are used as its terminals, i.e., current enters at A and leaves at B (C is left isolated). The value of total current I is

A. 1.1 A

B. 100 mA

C. 900 mA

D. 1 A

Answer : Option D
Explaination / Solution:

Thee above circuit can be redrawn as:

Total resistance in the arm EF ( milliammeter and the 0.9Ω resistor)= 9+0.9=9.9Ω.

Since the milliammeter gives full scale deflection when A and B are used as its terminals, the current in the arm EF is 10 mA. The potential difference across the arm is

VEF = Ig* R = 10(10^ - 3) * 9.9 = 0.099V.

The potential difference across AB= potential diff across the 0.1Ω resistor= VEF = 0.099 V.

The current through the 0.1Ω resistor

$\begin{array}{l} I_{0.1} = \frac{V_{E F}}{0.1} = \frac{0.099}{0.1} \\ I_{0.1} = 0.99 A \end{array}$

The total current I is the sum of I0.1 and Ig

= Ig + I0.1 = 0.01 + 0.99 = 1A.

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q3. Current density of a conductor is

A. the net charge flowing through the area

B. the net current flowing through the area normally per unit time

C. Is always zero

D. the net charge flowing through the area per unit time

Answer : Option B
Explaination / Solution:

Current density J = I/A In electromagnetism, current density is the electric current per unit area of cross section. It is a vector and has a direction along the area vector.

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q4. Mobility is defined as

A. the magnitude of the drift velocity per unit voltage

B. the magnitude of the drift velocity per unit charge

C. the number of charges in motion per unit electric field

D. The magnitude of the drift velocity per unit electric field.

Answer : Option D
Explaination / Solution:

Mobility is defined as the drift velocity acquired by the charge per unit electric field strength. Faster the particle moves in a given electric field strength greater is the mobility.

$\begin{array}{l} m_{μ} = \frac{V_{d}}{E} \end{array}$

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q5. The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. The current flowing in the potentiometer is:

A. 0.8 A

B. 0.4 A

C. 0.2 A

D. 0.1 A

Answer : Option B
Explaination / Solution:

If the battery has an e.m.f E, resitance of the potentiometer is R and the internal resistance of the battery is r, then the current I flowing in the potentiometer wire is given as

$\begin{array}{l} I = \frac{E}{(R + r)} \\ I = \frac{2}{(4 + 1)} \\ I = 0.4 A \end{array}$

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q6. The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. The p.d. across the potentiometer wire is:

A. 1.2 V

B. 1.6 V

C. 0.8 V

D. 2.0 V

Answer : Option B
Explaination / Solution:

If the battery has an e.m.f E, the resitance of the potentiometer is R and internal resistance of the battery is r, then the current I flowing in the potentiometer wire is given as,

$\begin{array}{l} I = \frac{E}{(R + r)} \\ I = \frac{2}{(4 + 1)} \\ I = 0.4 A \end{array}$

The potential difference V across the potentiometer is

$\begin{array}{l} V = I \times R \\ \Rightarrow V = 0.4 \times 4 \\ V = 1.6 V \end{array}$

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q7. The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm, the potential gradient in the potentiometer wire is:

A. 0.8 volt/m

B. 2.0 volt/m

C. 1.2 volt/m

D. 1.6 volt/m

Answer : Option D
Explaination / Solution:

If the battery has an e.m.f E, resitance of the potentiometer is R and the internal resistance of the battery is r, then the current I flowing in the potentiometer wire is given as,

$\begin{array}{l} I = \frac{E}{(R + r)} \\ I = \frac{2}{(4 + 1)} \\ I = 0.4 A \end{array}$

The potential difference V across the potentiometer,

$\begin{array}{l} V = I \times R \\ \Rightarrow V = 0.4 \times 4 \\ V = 1.6 V \end{array}$

The potential gradient = ( potential drop across the potentiometer)/ length of the potentiometer wire)

= V/l

= 1.6/1

Potential gradient = 1.6 V/m

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q8. The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm, if a cell of e.m.f. 1.2 volt is balanced by it, the balancing length will be

A. 50 cm

B. 90 cm

C. 60 cm

D. 75 cm

Answer : Option D
Explaination / Solution:

If the battery has e.m.f E, resitance of the potentiometer is R and the internal resistance of the battery is r, then the current I flowing in the potentiometer wire is given by,

$\begin{array}{l} I = \frac{E}{(R + r)} \\ I = \frac{2}{(4 + 1)} \\ I = 0.4 A \end{array}$

The potential difference V across the potentiometer

$\begin{array}{l} V = I \times R \\ \Rightarrow V = 0.4 \times 4 \\ V = 1.6 V \end{array}$

The potential gradient = ( potential drop across the potentiometer)/ length of the potentiometer wire)

= V/l

= 1.6/1

Potential gradient = 1.6 V/m

The emf of the cell

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q9. A potentiometer has a uniform wire of length 10m and resistance 5 ohms. The potentiometer is connected to an external battery of emf of 10V and negligible internal resistance and a resistance of 995 ohms in series. The potential gradient along the wire is:

A. 1 mV/m

B. 1 mV/cm

C. 5 mV/cm

D. 5 mV/m

Answer : Option D
Explaination / Solution:

The total resistance is the sum of the resistance of the potentiometer and the external resistance.

R = Rpot + Rext = 5 + 995 = 1000 ohms .

The current through the potentiometer wire I = E/R = 10/1000 = 0.01A.

The potential drop across the potentiometer wire is

$\begin{array}{l} V = I \times R_{p o t} \\ \Rightarrow V = 0.01 \times 5 \\ V = 0.05 V \end{array}$

The potential gradient = ( potential drop across the potentiometer wire)/ length of the potentiometer wire)

= 0.05/10

= 5 × 10^-3 V/m

= 5 m V/m

# Current Electricity # CBSE 12th Physics Prepare / Learn

Q10. A potentiometer wire, 10 m long, has resistance 40 ohms. It is put in series with a resistance 760 ohms and connected to a 2 volt battery. The potential gradient in the wire is :

A. 1×10−2volt/m

B. 1×10−6volt/m

C. 1×10−3volt/m

D. 1×10−4volt/m

Answer : Option A
Explaination / Solution:

The total resistance is the sum of the resistance of the potentiometer and the external resistance. $R = R_{p o t} + R_{e x t} = 40 + 760 = 800 Ω$

Current through the potentiometer is

$\begin{array}{l} I = \frac{E}{R} = \frac{2}{800} \\ I = 2.5 \times 10^{- 3} A \end{array}$

The potential drop across the potentiometer

$\begin{array}{l} V = I \times R_{p o t} \\ V = (2.5 \times 10^{- 3}) \times 40 \\ \Rightarrow V = 0.1 V \end{array}$

The potential gradient = ( potential drop across the potentiometer)/ length of the potentiometer wire)

$\begin{array}{l} = \frac{0.1}{10} V / m \\ ∴ P o t e n t i a l g r a d i e n t = 1 \times 10^{- 2} V / m \end{array}$

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