Question

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the potential difference across the capacitor when I=2.0mA? At what rate is energy being stored in the capacitor when I = 2.0mA?

Answer 1

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

Answer 2

The potential difference across the capacitor is 30V and the rate is energy being stored in the capacitor is 0.06W

I = 2 mA and R = 10000 ohm, V = 50V, C=30µF

Hence voltage across resistor ($V_R$):

$V_R$ = IR = 2 * 10⁻³ * 10000 = 20V

The voltage across capacitor ($V_C$) is:

$V_C+V_R=V\\\\V_C+20=50\\\\V_C=30V$

The energy (U) stored in a capacitor is:

$U=\frac{1}{2}\frac{q^2}{C} \\\\\frac{dU}{dt} =\frac{q}{C} *\frac{dq}{dt}\\\\\frac{dU}{dt} =30 *2*10^{-3} =0.06W$

The potential difference across the capacitor is 30V and the rate is energy being stored in the capacitor is 0.06W

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