Question

four identical capacitors are connected as shown in diagram when a battery of 6 V is connected between A and B the charge stored is found to be 1.5 uC the value of C1 is

Answer 1

0.1μF is the value of $C_{1}$

Explanation:

Given that,

Voltage of battery = 6 V

So,

Capacitance through A and B = $\frac{C_{1} }{2} + C_{1} + C_{1}$

= $\frac{5}{2} C_{1}$

Since Q = CV

⇒ 1.5 μC = $\frac{5}{2} C_{1}$ * 6

⇒ $C_{1}$ = 1.5 * 10^-6

= 0.1 * 10^-6F

= 0.1μF

∵ $C_{1}$. = 0.1μF

Thus, 0.1μF is the value of $C_{1}$.

Learn more: Capacitor

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