A circuit has a section AB shown in the figure- The emf of the source equals - -10V- the capacitance as of the capacitors are equal to C1-1-0- F and C2-2-0- F- the potential difference -A-B-5-0V- The voltage across capacitor C1 C2 is respectively-

The correct option is A 10 V, 5 V Give, e= 10V #160; C _ 1 =1μ F C _ 2 =2μ F V _ A V _ B =5V ∴ by emf formula, V _ A V _ B ...

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Byju's Answer

Standard VIII

Physics

Capacitors

A circuit has...

Question

A circuit has a section AB shown in the figure. The emf of the source equals $ε = 10$ V, the capacitance as of the capacitors are equal to $C_{1} = 1.0 μ F$ and $C_{2} = 2.0 μ F$ , the potential difference $ϕ_{A} - ϕ_{B} = 5.0$ V. The voltage across capacitor $C_{1}$ & $C_{2}$ is respectively.

10 V, 5 V

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10 / 3 V, 10 / 3 V

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5 / 3 V, 5 / 3 V

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0 V, 0 V

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Solution

The correct option is A $10 V, 5 V$
Give, e= 10V
$C_{1} = 1 μ F C_{2} = 2 μ F$
$V_{A} - V_{B} = 5 V$
$∴$ by emf formula,
$V_{A} - V_{B} = \frac{q}{C_{1}} - e + \frac{q}{C_{2}}$
$V_{A} - V_{B} + e = \frac{q (C_{1} + C_{2})}{C_{1} C_{2}}$
$q = \frac{[(V_{A} - V_{B}) + e] C_{1} C_{2}}{C_{1} + C_{2}}$
Voltage across $C_{1}$ is $V_{1} = \frac{q}{C_{1}} = \frac{[(V_{A} - V_{B}) + e] C_{2}}{C_{1} + C_{2}}$
= $\frac{(5 + 10) \times 2}{1 + 2} = 10 V$
Voltage across $C_{2}$ is, $V_{2} = \frac{q}{C_{2}} = \frac{[(V_{A} - V_{B}) + e] C_{1}}{C_{1} + C_{2}}$
= $\frac{(5 + 10) \times 1}{1 + 2} = 5 V$

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A circuit has a section AB shown in the figure. The emf of the source equals ε=10V, the capacitance as of the capacitors are equal to C1=1.0μF and C2=2.0μF, the potential difference ϕA−ϕB=5.0V. The voltage across capacitor C1 & C2 is respectively.

The correct option is A 10V,5VGive, e= 10V C1=1μFC2=2μFVA−VB=5V∴ by emf formula,VA−VB=qC1−e+qC2VA−VB+e=q(C1+C2)C1C2q=[(VA−VB)+e]C1C2C1+C2 Voltage across C1 is V1=qC1=[(VA−VB)+e]C2C1+C2 =(5+10)×21+2=10VVoltage across C2 is, V2=qC2=[(VA−VB)+e]C1C1+C2 =(5+10)×11+2=5V

A circuit has a section AB shown in the figure. The emf of the source equals $ε = 10$ V, the capacitance as of the capacitors are equal to $C_{1} = 1.0 μ F$ and $C_{2} = 2.0 μ F$ , the potential difference $ϕ_{A} - ϕ_{B} = 5.0$ V. The voltage across capacitor $C_{1}$ & $C_{2}$ is respectively.