SATHEE: Current Electricity 1 Question 6

Current Electricity 1 Question 6

6. Determine the charge on the capacitor in the following circuit

(2019 Main, 9 April I)

(a) $2 μ C$

(b) $200 μ C$

(d) $60 μ C$

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Solution:

Given circuit is

To find charge on capacitor, we need to determine voltage across it. In steady state, capacitor will acts as open circuit and circuit can be reduced as

In series, $R_{eq} = 2 Ω + 10 Ω = 12 Ω$

$⇓$

In parallel, $R_{e q} = \frac{4 \times 12}{4 + 12} = 3 Ω$

In series, $R_{e q} = 6 Ω + 3 Ω = 9 Ω$

So, current in steady state, $I = \frac{V}{R} = \frac{72}{9} = 8 A$

Now, by using current division, at point $P$ , current in $6 Ω$ branch is

Current in $4 Ω$ branch is,

${\dot{I}}_{2} = \frac{V_{P} - 0}{4} = \frac{24 - 0}{4} = 6 Ω$

So, current in $2 Ω$ resistance is

$\begin{aligned} I_{1} & = 8 - I_{2} [∵ I = I_{1} + I_{2}] \\ = 8 - 6 = 2 A \end{aligned}$

$∴$ Potential difference across $10 Ω$ resistor is

$V_{Q G} = 2 A \times 10 Ω = 20 V$

Same potential difference will be applicable over the capacitor (parallel combination).

So, charge stored in the capacitor will be

$\begin{aligned} Q = C V = 10 \times 10^{- 6} \times 20 \\ \Rightarrow & Q = 2 \times 10^{- 4} C = 200 μ C \end{aligned}$

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अगला

Current Electricity 1 Question 6

6. Determine the charge on the capacitor in the following circuit

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

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सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Current Electricity 1 Question 6

6. Determine the charge on the capacitor in the following circuit

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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