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Standard XII

Physics

Series and Parallel Combination of Resistors

In the given ...

Question

In the given circuit, what is the voltage across capacitor when source voltage $ε$ is 12 V?

-8 V

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8 V

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-16 V

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16 V

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Solution

The correct option is D 16 V
$Z = X_{c} - X_{L}$ where $X_{C} = \frac{1}{ω c} = 8 Ω$ & $X_{L} = ω L = 2 Ω$

$Z = 6 Ω i_{0} = \frac{24}{6} = 4 A$

$V_{o c} = i_{0} X_{C} = 4 \times 8 = 32$ V

from phases when $ε = \frac{ε_{0}}{2}$

$V_{C} = \frac{V_{o c}}{2} = 16$ V

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In the given circuit, what is the voltage across capacitor when source voltage ε is 12 V?

The correct option is D 16 VZ=Xc−XL where XC=1ω c=8 Ω & XL=ω L=2 Ω Z=6 Ω i0=246=4 A Voc=i0XC=4×8=32 V from phases when ε=ε02 VC=Voc2=16 V

In the given circuit, what is the voltage across capacitor when source voltage $ε$ is 12 V?

The correct option is D 16 V
$Z = X_{c} - X_{L}$ where $X_{C} = \frac{1}{ω c} = 8 Ω$ & $X_{L} = ω L = 2 Ω$

$Z = 6 Ω i_{0} = \frac{24}{6} = 4 A$

$V_{o c} = i_{0} X_{C} = 4 \times 8 = 32$ V

from phases when $ε = \frac{ε_{0}}{2}$

$V_{C} = \frac{V_{o c}}{2} = 16$ V