In the post dated ^{th} June 2008

**(1)** In the circuit shown, the battery and the inductor have negligible resistance. The currents through R_{1}, R_{2} and R_{3} are i_{1}, i_{2} and i_{3} 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 R_{2 }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 R_{2} 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 R_{2}? Justify your answer.

(d) If a 16 Ω resistor is used instead of the capacitor mentioned in (b), calculate the current through R_{2} immediately after closing the switch S.

(e) Show the nature of variation of the potential difference across R_{2} 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 = I _{0}*(1– e

^{– Rt /L})

**where**

*I*

_{0}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 R_{2 }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 = V _{0}*(1– e

^{–t/RC})

**where**

*V*

_{0}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 R_{2 }immediately after closing the switch S is equal to the charging voltage *V*_{0} which is equal to the voltage across R_{3} given by

*V*_{0} = [12 V/(8+8) Ω] ×8 Ω = 6 V.

(c) After reaching steady state in case (b), the potential difference across the resistor R_{2} will be zero since the capacitor will be fully charged and there will be no charging current through R_{2} to produce a voltage drop across it.

(d) The current through R_{2 }will rise to the final steady value abruptly if a resistor is used instead of the capacitor mentioned in (b). The current through R_{2} immediately after closing the switch S is therefore equal to the current resulting from the division of the main current i_{1}.

Now, i_{1} = 12 V/(8+6) Ω = 6/7 A.

[The 6 Ω resistance substituted above is the parallel combined value of R_{3} = **8 Ω** and (R_{2} + 16 Ω) = **24 Ω**].

The above current gets divided between R_{3 }and the (R_{2} + 16 Ω) branch.

_{2}and 16 Ω = Main current×(Resistance of the other branch/Total resistance) = (6/7)×[8/(24+8)] = (3/14) A.

(e) As the potential difference across the capacitor is given by *V = V _{0}*(1– e

^{–t/RC}) where

*V*

_{0}= 6 volts, the potential difference across the resistor R

_{2}is

*V*

_{0 }–

*V*(1– e

_{0}^{–t/RC}) =

*V*e

_{0}^{–t/RC}.

[This follows from Kirchoff’s voltage law: *V*_{C} + *V*_{R2} = *V*_{0}]

The nature of variation of the potential difference across R_{2} 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 R_{2} is (3/14) A as already shown. The potential difference across R_{2} in this case is (3/14)×8 volts = 1.7 volt, nearly. As the current through R_{2} will rise abruptly from zero, the potential difference across R_{2} 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 i_{2} and i_{3} on opening the switch S? Put a tick mark against the correct option among the following:

No change____

Current i_{2} alone is reversed____

Current i_{3} alone is reversed____

Both i_{2} and i_{3} are reversed____

Justify your answer.

(b) Calculate the potential difference across resistance R_{2} immediately after opening switch S

(c) Calculate the total amount of heat generated in resistance R_{2} after switch S is *opened*.

(d) Calculate the *power* dissipated in resistance R_{2} 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 i_{2} will be unchanged. But the direction of i_{3} 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 R_{1} since the total resistance across the battery is 8+4 = 12 Ω. [Here 4 Ω is the parallel combined value of R_{2} and R_{3}]. The potential difference across R_{2} (and R_{3}) is therefore 4V. The current in the inductance branch cannot change immediately on opening the switch S so that the potential drop across R_{2} will be 4 volt itself immediately after opening the switch S.

(c) Since the voltage across R_{2 }is 4 V under steady state with the switch S closed, the current through R_{2} is 4 V/8 Ω = 0.5 A.The switch is opened at *t =0 *and the current (*I*_{0}) at* t* = 0 is 0.5 A itself. The current (*I *) decay*s* in accordance with the equation,

*I* = *I*_{0 }e^{– R t /L} where R = R_{2}+R_{3} = 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 R_{2 }= *I *^{2}R_{2}

= 0.25 e^{– 16 t }×8 = 2 e^{– 16 t}

Therefore, total heat (H) produced in R_{2} 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 R_{2} is 0.5 A as shown above. The power (P) dissipated in the resistance R_{2} is given by

P = I^{2}R_{2} = 0.5^{2}×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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