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Resistances R1 , R, R3 and R4 are connected as shown in the figure. S1 and Sare two keys. Discuss the current flowing in the circuit in the following cases.

(i)   Both S1and S2are closed.

(ii)  Both S1 and S2 are open.

(iii) S1 is closed but Sis open.

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(i) The circuit for this case is shown below.

R4 is shunted by S. Hence the Resistance of this combination will be practically zero

R1 and R2 in parallel

∴ \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)

∴ \(R_p = \frac{R_1 R_2}{R_1 + R_2}\)

R3 and Rp in series

∴ Rs = R3 + Rp

∴ I3 = \(\frac{V}{R_s}\) = \(\frac{V}{R_3 + R_p}\)

V1= V-I3R3

= V- \(\frac{R_3 V}{R_3 + R_p}\)

= V \((1-\frac{R_3}{R_3 + R_p})\)

= V\((\frac{R_p}{R_3 + R_p})\)

∴ I1= V1/R1 = \(\frac{V}{R_1}(\frac{R_p}{R_3 + R_p})\)

Similarly

I2 = \(\frac{V}{R_2}(\frac{R_p}{R_3 + R_p})\)

 

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(ii)  Both S1 and S2 are open.

Answer

The circuit for this case is shown below.

Here, R1R4 and R3 are in series. Therefore, the equivalent resistance of the circuit is

Req=R1+R3+R4

Hence, the current flowing in the circuit is

I= \(\frac{V}{R_(eq)}\) =  \(\frac{V}{R_1+R_3+R_4}\)

where, V is the potential difference of the battery.

 

 

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(iii) S1 is closed but Sis open.

 

The circuit for this case is shown below.

Here, R1 and R2 are in parallel. Their effective resistance is
R'=\(\frac{R_1×R_2}{R_1+R_2}\)

Now, R3R' and R4 are in series.  Thus, the equivalent resistance of the circuit is
Req= \(\frac{R_1×R_2}{R_1+R_2} + R_3 + R_4 \)

= \(\frac{R_1 R_2 +R_3 R_1 + R_3 R_2 + R_4 R_1 +R_4 R_2}{R_1+R_2}\)

Hence, the current flowing in the circuit is
I=V/Req= \(\frac{v}{\frac{R_1 R_2 +R_3 R_1 + R_3 R_2 + R_4 R_1 +R_4 R_2}{R_1+R_2}}\)

= \(\frac{V(R_1+R_2)}{R_1 R_2 +R_3 R_1 + R_3 R_2 + R_4 R_1 +R_4 R_2}\)

where, V is the potential difference of the battery.

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