In the circuit shown, charge on the 5 \mu F capacitor is :

Capacitor PYQ — JEE Main 2020 · 5 Sep · Shift 2 · Q57 with Solution

View written solutionFree

Correct answer: D

Let the two battery terminals be the left top node and left bottom node, with a potential difference of $20\text{ V}$ across the capacitor network (as shown in the circuit).

We need the charge on the $5\,\mu\text{F}$ capacitor.

1. Identify the capacitor combinations

From the figure, the network can be reduced as follows:

The $10\,\mu\text{F}$ and $5\,\mu\text{F}$ capacitors are in series.
Their equivalent is in parallel with the $4\,\mu\text{F}$ capacitor.
That entire combination is then in series with the $3\,\mu\text{F}$ capacitor across the battery.

2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

For series capacitors,

C_{s} = \frac{C_1 C_2}{C_1 + C_2}

So,

C_s = \frac{10 \times 5}{10+5} = \frac{50}{15} = \frac{10}{3}\,\mu\text{F}

3. Parallel combination with $4\,\mu\text{F}$

Now this is in parallel with $4\,\mu\text{F}$ :

C_p = 4 + \frac{10}{3} = \frac{12+10}{3} = \frac{22}{3}\,\mu\text{F}

4. Series combination with $3\,\mu\text{F}$

Now $C_p = \dfrac{22}{3}\,\mu\text{F}$ is in series with $3\,\mu\text{F}$ .

Hence equivalent capacitance of the whole circuit:

C_{eq} = \frac{\left(\frac{22}{3}\right)(3)}{\frac{22}{3}+3} = \frac{22}{\frac{31}{3}} = \frac{66}{31}\,\mu\text{F}

5. Total charge supplied by the battery

Using $Q = C_{eq}V$ with $V=20\text{ V}$ ,

Q = \frac{66}{31} \times 20 = \frac{1320}{31}\,\mu\text{C}

This is the charge on each element of the final series combination, i.e. on the $3\,\mu\text{F}$ capacitor and on the parallel block.

So charge on the parallel block is

Q_p = \frac{1320}{31}\,\mu\text{C}

6. Voltage across the parallel block

V_p = \frac{Q_p}{C_p} = \frac{\frac{1320}{31}}{\frac{22}{3}} = \frac{1320}{31}\cdot \frac{3}{22} = \frac{180}{31}\text{ V}

Thus the series pair $(10\,\mu\text{F}, 5\,\mu\text{F})$ has voltage

V = \frac{180}{31}\text{ V}

7. Charge on the $5\,\mu\text{F}$ capacitor

Since $10\,\mu\text{F}$ and $5\,\mu\text{F}$ are in series, both carry the same charge.

Charge on the series pair:

Q_{series} = C_s V = \frac{10}{3} \cdot \frac{180}{31} = \frac{600}{31}\,\mu\text{C}

Q_{series} \approx 19.35\,\mu\text{C}

Therefore, charge on the $5\,\mu\text{F}$ capacitor is

\boxed{19.35\,\mu\text{C}}

8. Compare with options

Given options are:

A: $5.45\,\mu\text{C}$
B: $18.00\,\mu\text{C}$
C: $10.90\,\mu\text{C}$
D: $16.36\,\mu\text{C}$

Our derived value $19.35\,\mu\text{C}$ does not match any option exactly.

Among the listed options, none is correct for the circuit reduction used above.

9. Comparison with stored correct answer

Stored correct answer: D ( $16.36\,\mu\text{C}$ )

Our derived answer does not agree with this.

This suggests either:

the circuit image has a different connection than inferred from the text, or
the stored answer is incorrect.

Based on the capacitor reduction described above, the charge on the $5\,\mu\text{F}$ capacitor should be

\boxed{19.35\,\mu\text{C}}

More from Capacitor

View written solutionFree

Correct answer: D

Let the two battery terminals be the left top node and left bottom node, with a potential difference of $20\text{ V}$ across the capacitor network (as shown in the circuit).

We need the charge on the $5\,\mu\text{F}$ capacitor.

1. Identify the capacitor combinations

From the figure, the network can be reduced as follows:

The $10\,\mu\text{F}$ and $5\,\mu\text{F}$ capacitors are in series.
Their equivalent is in parallel with the $4\,\mu\text{F}$ capacitor.
That entire combination is then in series with the $3\,\mu\text{F}$ capacitor across the battery.

2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

For series capacitors,

C_{s} = \frac{C_1 C_2}{C_1 + C_2}

So,

C_s = \frac{10 \times 5}{10+5} = \frac{50}{15} = \frac{10}{3}\,\mu\text{F}

3. Parallel combination with $4\,\mu\text{F}$

Now this is in parallel with $4\,\mu\text{F}$ :

C_p = 4 + \frac{10}{3} = \frac{12+10}{3} = \frac{22}{3}\,\mu\text{F}

4. Series combination with $3\,\mu\text{F}$

Now $C_p = \dfrac{22}{3}\,\mu\text{F}$ is in series with $3\,\mu\text{F}$ .

Hence equivalent capacitance of the whole circuit:

C_{eq} = \frac{\left(\frac{22}{3}\right)(3)}{\frac{22}{3}+3} = \frac{22}{\frac{31}{3}} = \frac{66}{31}\,\mu\text{F}

5. Total charge supplied by the battery

Using $Q = C_{eq}V$ with $V=20\text{ V}$ ,

Q = \frac{66}{31} \times 20 = \frac{1320}{31}\,\mu\text{C}

This is the charge on each element of the final series combination, i.e. on the $3\,\mu\text{F}$ capacitor and on the parallel block.

So charge on the parallel block is

Q_p = \frac{1320}{31}\,\mu\text{C}

6. Voltage across the parallel block

V_p = \frac{Q_p}{C_p} = \frac{\frac{1320}{31}}{\frac{22}{3}} = \frac{1320}{31}\cdot \frac{3}{22} = \frac{180}{31}\text{ V}

Thus the series pair $(10\,\mu\text{F}, 5\,\mu\text{F})$ has voltage

V = \frac{180}{31}\text{ V}

7. Charge on the $5\,\mu\text{F}$ capacitor

Since $10\,\mu\text{F}$ and $5\,\mu\text{F}$ are in series, both carry the same charge.

Charge on the series pair:

Q_{series} = C_s V = \frac{10}{3} \cdot \frac{180}{31} = \frac{600}{31}\,\mu\text{C}

Q_{series} \approx 19.35\,\mu\text{C}

Therefore, charge on the $5\,\mu\text{F}$ capacitor is

\boxed{19.35\,\mu\text{C}}

8. Compare with options

Given options are:

A: $5.45\,\mu\text{C}$
B: $18.00\,\mu\text{C}$
C: $10.90\,\mu\text{C}$
D: $16.36\,\mu\text{C}$

Our derived value $19.35\,\mu\text{C}$ does not match any option exactly.

Among the listed options, none is correct for the circuit reduction used above.

9. Comparison with stored correct answer

Stored correct answer: D ( $16.36\,\mu\text{C}$ )

Our derived answer does not agree with this.

This suggests either:

the circuit image has a different connection than inferred from the text, or
the stored answer is incorrect.

Based on the capacitor reduction described above, the charge on the $5\,\mu\text{F}$ capacitor should be

\boxed{19.35\,\mu\text{C}}

Capacitor question

Capacitor question

1. Identify the capacitor combinations

2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

3. Parallel combination with $4\,\mu\text{F}$

4. Series combination with $3\,\mu\text{F}$

5. Total charge supplied by the battery

6. Voltage across the parallel block

7. Charge on the $5\,\mu\text{F}$ capacitor

8. Compare with options

9. Comparison with stored correct answer

More from Capacitor

Capacitor question

Capacitor question

1. Identify the capacitor combinations

2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

3. Parallel combination with $4\,\mu\text{F}$

4. Series combination with $3\,\mu\text{F}$

5. Total charge supplied by the battery

6. Voltage across the parallel block

7. Charge on the $5\,\mu\text{F}$ capacitor

8. Compare with options

9. Comparison with stored correct answer

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1. Identify the capacitor combinations

2. Equivalent of 10 μF10\,\mu\text{F}10μF and 5 μF5\,\mu\text{F}5μF in series

3. Parallel combination with 4 μF4\,\mu\text{F}4μF

4. Series combination with 3 μF3\,\mu\text{F}3μF

5. Total charge supplied by the battery

6. Voltage across the parallel block

7. Charge on the 5 μF5\,\mu\text{F}5μF capacitor

8. Compare with options

9. Comparison with stored correct answer

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1. Identify the capacitor combinations

2. Equivalent of 10 μF10\,\mu\text{F}10μF and 5 μF5\,\mu\text{F}5μF in series

3. Parallel combination with 4 μF4\,\mu\text{F}4μF

4. Series combination with 3 μF3\,\mu\text{F}3μF

5. Total charge supplied by the battery

6. Voltage across the parallel block

7. Charge on the 5 μF5\,\mu\text{F}5μF capacitor

8. Compare with options

9. Comparison with stored correct answer

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2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

3. Parallel combination with $4\,\mu\text{F}$

4. Series combination with $3\,\mu\text{F}$

7. Charge on the $5\,\mu\text{F}$ capacitor

2. Equivalent of $10\,\mu\text{F}$ and $5\,\mu\text{F}$ in series

3. Parallel combination with $4\,\mu\text{F}$

4. Series combination with $3\,\mu\text{F}$

7. Charge on the $5\,\mu\text{F}$ capacitor