Solution:

Here's the short solution for the circuit:

1. Just after the switch is closed:
   • The capacitors act as open circuits because they are initially uncharged (capacitor voltage cannot change instantaneously).
   • This means no current flows through the branches containing the capacitors.
2. Current Distribution:
   • The total resistance in the circuit is only the sum of the resistors in the main loop (2Ω + 8Ω), as the branches with capacitors are effectively open.
   • Total resistance = $2\Omega + 8\Omega = 10\Omega$.
   • Current $I_1 = \frac{\text{Voltage}}{\text{Resistance}} = \frac{6V}{2\Omega + 8\Omega} = 0.6A$.
3. Branch currents:
   • $I_3 = 0$ since the capacitor branch is open.
   • $I_2 = 0$ for the same reason as above.

Final Answer:

• \( I_1 = 0.6 A, \ I_2 = 0\)

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Solution:

At t =0 capacitor behaves as closed circuit to the 1000 ohm resistor connected in parallel with capacitor will get short circuited.

current  through the other resistor = 2/1000 = 2mA

At t = infinite capacitor behaves as open circuit so equivalent resistance becomes R=1000+1000 = 2000 Ohm

current  through the  resistor = 2/2000 = 1 mA

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Solution:

In steady state no current flows through the capacitor so, current through 4 ohm resistor will be zero.

In absence of 4 ohm resistor, total resistance of circuit is (1.2+2.8)= 4 ohm.

Total current given by battery = 6/4=1.5 Ampere.

In parallel combination current divides in reverse ratio of resistors: (3/5)*1.5= 0.9 Ampere

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Solution:

The charging of a capacitor through a resistor is described by the following equation:

$$Q(t) = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Where:

• $Q(t)$ is the charge on the capacitor at time $t$,
• $Q_{\text{max}}$ is the maximum (steady-state) charge the capacitor can hold,
• $\tau$ is the time constant, $\tau = R \cdot C$, where $R$ is the resistance and $C$ is the capacitance,
• $t$ is the time.

Step 1: Given condition (10% of steady charge)

We are told that at time $t$, the capacitor has collected 10% of the steady charge, so:

$$Q(t) = 0.1 \cdot Q_{\text{max}}$$

Step 2: Substitute into the charging equation

Substitute $Q(t) = 0.1 \cdot Q_{\text{max}}$ into the charging formula:

$$0.1 \cdot Q_{\text{max}} = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Cancel $Q_{\text{max}}$ from both sides:

$$0.1 = 1 - e^{-t/\tau}$$

Step 3: Solve for $t$

Rearrange the equation to solve for $e^{-t/\tau}$:

$$e^{-t/\tau} = 1 - 0.1 = 0.9$$

Take the natural logarithm of both sides:

$$-\frac{t}{\tau} = \ln(0.9)$$ $$t = -\tau \ln(0.9)$$

Using the fact that $\ln(0.9) = -\ln(10/9)$:

$$t = \tau \ln\left(\frac{10}{9}\right)$$

Final Answer:

The time at which the capacitor has collected 10% of the steady charge is $t = \tau \ln\left(\frac{10}{9}\right)$.

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