Calculate the potential difference across the capacitor in the circuit (in Volts). (Approximate the answer to the nearest integer)

Question

Calculate the potential difference across the capacitor in the circuit (in Volts). (Approximate the answer to the nearest integer) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-6aethscta95720ca.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-6aethscta957ialn.jpg" /> Net current at point x=0. ∴ (0 - x/10)+(8 - x/4)+(0 - (x + 2)/5)=0⇒ x=2.909 Potential across capacitor ⇒ [(x + 2) - 8]=-3.09≃ 3

Q. Calculate the potential difference across the capacitor in the circuit (in Volts). (Approximate the answer to the nearest integer)

Net current at point $x = 0.$
$∴ \frac{0 - x}{10} + \frac{8 - x}{4} + \frac{0 - ( x + 2 )}{5} = 0 \Rightarrow x = 2.909$
Potential across capacitor $\Rightarrow [(x + 2) - 8] = - 3.09 ≃ 3$

Q. Calculate the potential difference across the capacitor in the circuit (in Volts). (Approximate the answer to the nearest integer)

Net current at point x=0. ∴100−x​+48−x​+50−(x+2)​=0⇒x=2.909 Potential across capacitor ⇒[(x+2)−8]=−3.09≃3

Net current at point $x = 0.$
$∴ \frac{0 - x}{10} + \frac{8 - x}{4} + \frac{0 - ( x + 2 )}{5} = 0 \Rightarrow x = 2.909$
Potential across capacitor $\Rightarrow [(x + 2) - 8] = - 3.09 ≃ 3$