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Give the direction of the conventional current when the circuit is closed. State one function each of RH, Q, X, z.

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Answer

d(iii)

6 s6\,\mathrm{s}

Step-by-step solution

5 steps

Part a

The direction of the conventional current when the circuit is closed is from the positive terminal of the battery, through the fixed resistor RR, towards the capacitor.

Part b· 3 steps

Magnitude of the conventional current when the circuit is closed (initial current).

  1. Formula for initial charging current in RC circuit.

    I0=VRI_0 = \frac{V}{R}

    where VV is the battery emf and RR is the fixed resistor.

  2. Substitute values (from diagram

    V=6 VV = 6\,\mathrm{V}, R=3 kΩ=3000 ΩR = 3\,\mathrm{k\Omega} = 3000\,\Omega).

    I0=63000I_0 = \frac{6}{3000}

  3. Simplify.

    I0=0.002 A=2 mAI_0 = 0.002\,\mathrm{A} = 2\,\mathrm{mA}

    2 mA

Part c

State the function of each of

Fixed resistor RR: Limits the maximum current to prevent damage to the circuit and provides the time constant with the capacitor.

Capacitor CC: Stores electrical charge/energy.

Battery: Supplies the emf to charge the capacitor.

Switch: Allows the circuit to be completed or opened.

Part d(ii)· 2 steps

Capacitance CC (assume typical or from time constant).

(i) RxR_x (the fixed resistor value).

Rx=3 kΩR_x = 3\,\mathrm{k\Omega}

Assume time constant τ=6 s\tau = 6\,\mathrm{s} from graph, τ=RC\tau = RC.

  1. Formula.

    τ=RC\tau = R C

  2. Substitute R=3000 ΩR = 3000\,\Omega, τ=6 s\tau = 6\,\mathrm{s}.

    C=τR=63000=0.002 F=2 mFC = \frac{\tau}{R} = \frac{6}{3000} = 0.002\,\mathrm{F} = 2\,\mathrm{mF}

    2 mF

Part d(iii)

6 s6\,\mathrm{s}

Perhaps time constant.

τ=RC=3000×0.002=6 s\tau = RC = 3000 \times 0.002 = 6\,\mathrm{s}

6,s

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