NEET Current Electricity MCQs & PYQs

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

Current through each resistor is 220/1100 = 1/5 A Total current through ammeter = 3 * 1/5 A= 3/5 A

Q: 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:

To find the drift velocity vdv_d of the electrons, we use the formula for current in terms of drift velocity: I=nAevdI = n A e v_d Where: II is the current (1.2 A), nn is the number of free electrons per unit volume ( 5×1028 m−35 \times 10^{28} \, \text{m}^{-3} ), AA is the cross-sectional area of the conductor ( 10−4 m210^{-4} \, \text{m}^2 ), ee is the charge of an electron ( 1.6×10−19 C1.6 \times 10^{-19} \, \text{C} ), vdv_d is the drift velocity of the electrons (which we need to calculate). Step 1: Rearranging the formula to solve for vdv_d vd=InAev_d = \frac{I}{n A e} Step 2: Substituting the given values vd=1.2(5×1028)×(10−4)×(1.6×10−19)v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})} Step 3: Performing the calculation vd=1.2(5×1028)×(10−4)×(1.6×10−19)v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})} vd=1.28×106v_d = \frac{1.2}{8 \times 10^{6}} vd=1.5×10−7 m/sv_d = 1.5 \times 10^{-7} \, \text{m/s} Final Answer: The drift velocity of the electrons is 1.5×10−7 m/s\boxed{1.5 \times 10^{-7} \, \text{m/s}} .

Q: 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

Initial Setup: An ammeter and a voltmeter are connected in series to a cell. The ammeter reading is AA (current through the circuit). The voltmeter reading is VV (potential difference across the cell). What Happens When a Resistance is Added in Parallel with the Voltmeter? The voltmeter has a high internal resistance ( RvR_v ) to minimize current flow through it. When an additional resistance ( RR ) is connected in parallel with the voltmeter, the effective resistance of the voltmeter decreases because: Reff=Rv⋅RRv+R.R_{\text{eff}} = \frac{R_v \cdot R}{R_v + R}. Since RR is finite, Reff<RvR_{\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 ReffR_{\text{eff}} reduces the total resistance of the circuit. Lower resistance means higher total current through the circuit (Ohm's law: I=VRtotalI = \frac{V}{R_{\text{total}}} ). Thus, the ammeter reading ( AA ) 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 ( VV ) decreases. Conclusion: When a resistance is added in parallel with the voltmeter: The ammeter reading ( AA ) increases. The voltmeter reading ( VV ) decreases.

Practice NEET Current Electricity 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 :

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.

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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 :

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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:

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.

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Question 6:

moderate

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

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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

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.