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(1) In the circuit shown, the battery and the inductor have negligible resistance. The currents through R1, R2 and R3 are i1, i2 and i3 respectively when the switch S is kept closed. The switch S is closed at time t = 0.
(a) Calculate the potential difference across the 8 Ω resistor R2 immediately after closing the switch S.
(b) If the inductance is replaced by an uncharged 10 μF capacitor what will be the potential difference across the 8 Ω resistor R2 immediately after closing the switch S? Justify your answer.
(c) After reaching steady state, in case (b) what will be the potential difference across the resistor R2? Justify your answer.
(d) If a 16 Ω resistor is used instead of the capacitor mentioned in (b), calculate the current through R2 immediately after closing the switch S.
(e) Show the nature of variation of the potential difference across R2 with time t in cases (b) and (d).
(a) After closing the switch S, the current I in the inductance branch grows exponentially with time t in accordance with the equation,
I = I0(1– e– Rt /L) where I0 is the final steady current.
The current I is therefore zero when t = 0 so that the the potential difference across the 8 Ω resistor R2 immediately after closing the switch S is zero.
(b) After closing the switch S, the potential difference V across the capacitor (C) will build up exponentially with time t in accordance with the equation,
V = V0(1– e–t/RC) where V0 is the final steady voltage across the capacitor and R is the resistance in series with the capacitor.
The voltage V across the capacitor is therefore zero when t = 0 so that the potential difference across the 8 Ω resistor R2 immediately after closing the switch S is equal to the charging voltage V0 which is equal to the voltage across R3 given by
V0 = [12 V/(8+8) Ω] ×8 Ω = 6 V.
(c) After reaching steady state in case (b), the potential difference across the resistor R2 will be zero since the capacitor will be fully charged and there will be no charging current through R2 to produce a voltage drop across it.
(d) The current through R2 will rise to the final steady value abruptly if a resistor is used instead of the capacitor mentioned in (b). The current through R2 immediately after closing the switch S is therefore equal to the current resulting from the division of the main current i1.
Now, i1 = 12 V/(8+6) Ω = 6/7 A.
[The 6 Ω resistance substituted above is the parallel combined value of R3 = 8 Ω and (R2 + 16 Ω) = 24 Ω].
The above current gets divided between R3 and the (R2 + 16 Ω) branch.
(e) As the potential difference across the capacitor is given by V = V0(1– e–t/RC) where V0 = 6 volts, the potential difference across the resistor R2 is V0 – V0(1– e–t/RC) = V0 e–t/RC.
[This follows from Kirchoff’s voltage law: VC + VR2 = V0]
The nature of variation of the potential difference across R2 in this case (b) is as shown by curve (b).
When a 16 Ω resistor is used in place of the capacitor, the final current through R2 is (3/14) A as already shown. The potential difference across R2 in this case is (3/14)×8 volts = 1.7 volt, nearly. As the current through R2 will rise abruptly from zero, the potential difference across R2 will rise abruptly to 1.7 volt as shown by curve (d).
(2) In the above circuit, suppose the switch S was closed for sufficiently long time so that steady state was reached. In this steady state the.switch S is opened at time t = 0.
(a) Will there be any change in the directions of the currents i2 and i3 on opening the switch S? Put a tick mark against the correct option among the following:
No change____
Current i2 alone is reversed____
Current i3 alone is reversed____
Both i2 and i3 are reversed____
Justify your answer.
(b) Calculate the potential difference across resistance R2 immediately after opening switch S
(c) Calculate the total amount of heat generated in resistance R2 after switch S is opened.
(d) Calculate the power dissipated in resistance R2 in the steady state, before opening the switch S.
(a) On opening the switch S, the inductor will try to maintain the current flowing in it because of its inductance (electrical inertia). Therefore, the direction of i2 will be unchanged. But the direction of i3 will be reversed since the current now flows because of the voltage induced in the inductance.
(b) Under steady state, the battery delivers a current of 1 A through R1 since the total resistance across the battery is 8+4 = 12 Ω. [Here 4 Ω is the parallel combined value of R2 and R3]. The potential difference across R2 (and R3) is therefore 4V. The current in the inductance branch cannot change immediately on opening the switch S so that the potential drop across R2 will be 4 volt itself immediately after opening the switch S.
(c) Since the voltage across R2 is 4 V under steady state with the switch S closed, the current through R2 is 4 V/8 Ω = 0.5 A.The switch is opened at t =0 and the current (I0) at t = 0 is 0.5 A itself. The current (I ) decays in accordance with the equation,
I = I0 e– R t /L where R = R2+R3 = 8+8 =16 Ω (which is the total resistance in series with L, on opening the switch S).
Therefore, we have I = 0.5 e–16t/2 = 0.5 e– 8 t
Heat produced per second in the resistance R2 = I 2R2
= 0.25 e– 16 t ×8 = 2 e– 16 t
Therefore, total heat (H) produced in R2 after opening the switch S is given by
H = 0 ∫∞ 2e– 16 t = 0.125 joule.
(d) In the steady state before opening the switch S, the current flowing in R2 is 0.5 A as shown above. The power (P) dissipated in the resistance R2 is given by
P = I2R2 = 0.52×8 = 2 watt.
Note: Even if your theoretical steps, substitutions and calculations are correct, you will lose points if proper units are not supplied. This is especially important in questions involving numerical values.
Work out as many questions as possible to make you strong in facing the AP Physics Exam.
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