Two equal resistances,400 Omega each,are conn | vedclass

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(B) Let the initial resistances be $R_1 = 400 \Omega$ and $R_2 = 400 \Omega$. The total resistance is $R_{eq} = R_1 + R_2 = 800 \Omega$. The current i

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increase it by $2 \Omega$

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(B) Let the initial resistances be $R_1 = 400 \Omega$ and $R_2 = 400 \Omega$. The total resistance is $R_{eq} = R_1 + R_2 = 800 \Omega$. The current in the circuit is $I = V / R_{eq} = 8 / 800 = 0.01 A$. The potential difference across the second resistor is $V_2 = I R_2 = 0.01 	imes 400 = 4 V$.When $R_1$ increases by $0.5 \%$,the new resistance $R_1' = 400 + (0.5 / 100) 	imes 400 = 400 + 2 = 402 \Omega$.To keep the potential difference $V_2$ across $R_2$ constant at $4 V$,the current $I'$ must remain $0.01 A$. Since $I' = V / (R_1' + R_2')$,we have $0.01 = 8 / (402 + R_2')$.Solving for $R_2'$,we get $402 + R_2' = 8 / 0.01 = 800$,which gives $R_2' = 800 - 402 = 398 \Omega$.The change in $R_2$ is $\Delta R_2 = R_2' - R_2 = 398 - 400 = -2 \Omega$. Wait,let's re-evaluate: If $R_1$ increases,the total resistance increases,causing the current to decrease. To keep $V_2$ constant,$R_2$ must be adjusted. If $V_2 = I R_2 = (V / (R_1 + R_2)) R_2$,for $V_2$ to be constant,the ratio $R_2 / (R_1 + R_2)$ must be constant. This implies $R_2 / R_1$ must remain constant. Since $R_1$ increased by $0.5 \%$,$R_2$ must also increase by $0.5 \%$. Increase in $R_2 = 0.5 \% 	ext{ of } 400 = 2 \Omega$.

Two equal resistances,$400 \Omega$ each,are connected in series with an $8 V$ battery. If the resistance of the first one increases by $0.5 \%$,what is the change required in the resistance of the second one in order to keep the potential difference across it unaltered?

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