In the circuit shown,the reading of the ammet | vedclass

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(A) Case $1$: When both switches are open,the circuit consists of a $1.5 \, V$ battery in series with a $300 \, \Omega$ resistor,a $100 \, \Omega$ res

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$600 \, \Omega$

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(A) Case $1$: When both switches are open,the circuit consists of a $1.5 \, V$ battery in series with a $300 \, \Omega$ resistor,a $100 \, \Omega$ resistor,and a $50 \, \Omega$ resistor.The total resistance is $R_{eq1} = 300 + 100 + 50 = 450 \, \Omega$.The current through the ammeter is $I = \frac{1.5}{450} = \frac{1}{300} \, A$.Case $2$: When both switches are closed,the $R$ resistor is in parallel with the $100 \, \Omega$ resistor branch (containing the ammeter). Let the equivalent resistance of the parallel part be $R_p = \frac{100 R}{100 + R}$.The total circuit resistance is $R_{eq2} = 300 + R_p + 50 = 350 + \frac{100 R}{100 + R}$.The total current from the battery is $I_{total} = \frac{1.5}{R_{eq2}}$.The current through the ammeter branch is $I' = I_{total} 	imes \frac{R}{100 + R} = \frac{1.5}{350 + \frac{100 R}{100 + R}} 	imes \frac{R}{100 + R} = \frac{1.5 R}{350(100 + R) + 100 R} = \frac{1.5 R}{35000 + 450 R}$.Equating $I = I'$,we get $\frac{1}{300} = \frac{1.5 R}{35000 + 450 R}$.$35000 + 450 R = 450 R$. This suggests a re-evaluation of the circuit diagram. Based on the provided solution image,the $50 \, \Omega$ resistor is bypassed when switches are closed.Correct approach: With switches closed,the $50 \, \Omega$ resistor is shorted. The circuit is $1.5 \, V$ battery,$300 \, \Omega$ resistor,and $100 \, \Omega$ resistor in series with $R$ in parallel. $I = \frac{1.5}{300 + \frac{100 R}{100 + R}} 	imes \frac{R}{100 + R} = \frac{1.5 R}{300(100 + R) + 100 R} = \frac{1.5 R}{30000 + 400 R}$.Equating $\frac{1}{300} = \frac{1.5 R}{30000 + 400 R} \Rightarrow 30000 + 400 R = 450 R \Rightarrow 50 R = 30000 \Rightarrow R = 600 \, \Omega$.

In the circuit shown,the reading of the ammeter (ideal) is the same with both switches open as with both closed. Find the value of resistance $R$ in $\Omega$.

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