Given C1 - C5 - C3 - 2 - F- C2 - C4 - 4 - F- 2 - F can withstand maximum voltage of 5 V and 4 - F can withstand maximum of 10 V- Which of the capacitors will burn-

Q_1 = q_2 = 15 ×8/6⇒ V_1 = q_1/C_1 = 15 × 8/6 × 2 = 10 V (will burn) amp; V_2 = 5 V (will not burn) q_3 + q_4 = q_5 = 3/2 nbsp;× 15 = 45/2 V_5=q_5/C_5=45/2 ...

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Byju's Answer

Standard XI

Physics

Parallel and Series Combination of Capacitors

Given C1 = C5...

Question

Given $C_{1} = C_{5} = C_{3} = 2 μ F, C_{2} = C_{4} = 4 μ F$ .
$2 μ F$ can withstand maximum voltage of $5 V$ and $4 μ F$ can withstand maximum of $10 V$ . Which of the capacitors will burn?

All

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C_{1}

and

C_{2}

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C_{1}

and

C_{5}

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C_{1}, C_{2}

and

C_{5}

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Solution

The correct option is C $C_{1}$ and $C_{5}$
$q_{1} = q_{2} = 15 \times \frac{8}{6} \Rightarrow V_{1} = \frac{q_{1}}{C_{1}} = \frac{15 \times 8}{6 \times 2}$
$= 10 V$ (will burn)
& $V_{2} = 5 V$ (will not burn)
$q_{3} + q_{4} = q_{5} = \frac{3}{2} \times 15 = \frac{45}{2}$
$V_{5} = \frac{q_{5}}{C_{5}} = \frac{45}{2 \times 2} = 11.25 V$ (will burn)
$V_{3} = V_{4} = 15 - V_{5} = 3.75 V$ (will not burn)

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Given C1=C5=C3=2 μF, C2=C4=4 μF. 2 μF can withstand maximum voltage of 5 V and 4 μF can withstand maximum of 10 V. Which of the capacitors will burn?

The correct option is C C1 and C5q1=q2=15×86⇒V1=q1C1=15×86×2 =10 V (will burn) & V2=5 V (will not burn) q3+q4=q5=32×15=452 V5=q5C5=452×2=11.25 V(will burn) V3=V4=15−V5=3.75 V (will not burn)

Given $C_{1} = C_{5} = C_{3} = 2 μ F, C_{2} = C_{4} = 4 μ F$ .
$2 μ F$ can withstand maximum voltage of $5 V$ and $4 μ F$ can withstand maximum of $10 V$ . Which of the capacitors will burn?