Current Electricity

Practice Questions:

Question 1: moderate

Five identical resistors, each of value 1100Ω, are connected to a 220V battery as shown. The reading of ideal ammeter is :

1. \[\frac{1}{3} A\]

2. \[\frac{1}{5} A\]

3. \[\frac{3}{5} A\]

4. \[\frac{4}{5} A\]

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Current through each resistor is 220/1100 = 1/5 A

Total current through ammeter = 3 * 1/5 A= 3/5 A

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Question 2: moderate

In the circuit shown the reading of ammeter is 2A. The ammeter has negligible resistance. The value of R equals.

1. 2Ω

2. 4Ω

3. 6Ω

4. 8Ω

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Question 3: moderate

The V-I graph for two conductor of resistance R1 & R2 are as shown in the figure, the resistance of their series combination (R2 + R1) is proportional to :

1. /[\frac{1}{sin^{2}\theta cos^{2}\theta}/]

2. /[\frac{1}{sin\theta cos\theta}/]

3. /[\frac{cos\theta}{sin^{2}\theta}/]

4. /[\frac{tan2\theta}{sin^{2}\theta}/]

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Question 4: moderate

If the number of free electrons is $5\times 10^{28} m^{-3}$ then the drift velocity of electron in a conductor of area of cross-section $10^{-4} m^{2}$ for a current of 1.2 A is:

1. \[1.5\times 10^{-2} m/s\]

2. \[1.5\times 10^{-3} m/s\]

3. \[1.5\times 10^{-4} m/s\]

4. \[1.5\times 10^{-6} m/s\]

View Answer

To find the drift velocity

$v_d$

of the electrons, we use the formula for current in terms of drift velocity:

$I = n A e v_d$

Where:

$I$

is the current (1.2 A),
$n$

is the number of free electrons per unit volume ( $5 \times 10^{28} \, \text{m}^{-3}$

),
$A$

is the cross-sectional area of the conductor ( $10^{-4} \, \text{m}^2$

),
$e$

is the charge of an electron ( $1.6 \times 10^{-19} \, \text{C}$

),
$v_d$

is the drift velocity of the electrons (which we need to calculate).

Step 1: Rearranging the formula to solve for $v_d$

$v_d = \frac{I}{n A e}$

Step 2: Substituting the given values

$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$

Step 3: Performing the calculation

$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$

$v_d = \frac{1.2}{8 \times 10^{6}}$

$v_d = 1.5 \times 10^{-7} \, \text{m/s}$

Final Answer:

The drift velocity of the electrons is

$\boxed{1.5 \times 10^{-7} \, \text{m/s}}$

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Question 5: moderate

Find effective resistance between A & C if each resistor is of 6Ω.

1. 4 Ω

2. 2 Ω

3. 2.4 Ω

4. 3 Ω

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Question 6: moderate

Potential difference between C and B /[\left( V_{C}-V_{B} \right)/] in the circuit is :

1. 15 V

2. 38 V

3. 20 V

4. 9 V

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Question 7: moderate

An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter

1. both A and V will increase

2. both A and V will decrease

3. A will decrease, V will increase

4. A will increase, V will decrease

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Initial Setup:

An ammeter and a voltmeter are connected in series to a cell.
The ammeter reading is $A$

(current through the circuit).
The voltmeter reading is $V$

(potential difference across the cell).

What Happens When a Resistance is Added in Parallel with the Voltmeter?

The voltmeter has a high internal resistance ( $R_v$

) to minimize current flow through it.
When an additional resistance ( $R$

) is connected in parallel with the voltmeter, the effective resistance of the voltmeter decreases because: $R_{\text{eff}} = \frac{R_v \cdot R}{R_v + R}.$

Since $R$

is finite, $R_{\text{eff}} < R_v$

.

Effect on the Circuit:

Decrease in total resistance:
- The voltmeter (and its parallel combination) is in series with the ammeter.
- The decrease in $R_{\text{eff}}$
  
  reduces the total resistance of the circuit.
- Lower resistance means higher total current through the circuit (Ohm's law: $I = \frac{V}{R_{\text{total}}}$
  
  ).
- Thus, the ammeter reading ( $A$
  
  ) increases.
Voltage across the voltmeter decreases:
- With the reduced effective resistance of the voltmeter, a smaller fraction of the total voltage is dropped across it.
- Hence, the voltmeter reading ( $V$
  
  ) decreases.