Four resistances 40 Omega, 60 Omega, 90 Om | vedclass

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Question

(C) The circuit consists of two parallel branches connected across the $40 \ V$ battery. Branch $ABC$ has resistances $40 \ \Omega$ and $60 \ \Omega$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The circuit consists of two parallel branches connected across the $40 \ V$ battery. Branch $ABC$ has resistances $40 \ \Omega$ and $60 \ \Omega$ in series. Total resistance $R_1 = 40 + 60 = 100 \ \Omega$. Current $i_1 = \frac{V}{R_1} = \frac{40}{100} = 0.4 \ A$. Potential at $B$ relative to $A$ is $V_A - V_B = i_1 	imes 40 = 0.4 	imes 40 = 16 \ V$. Thus,$V_B = V_A - 16$. Branch $ADC$ has resistances $90 \ \Omega$ and $110 \ \Omega$ in series. Total resistance $R_2 = 90 + 110 = 200 \ \Omega$. Current $i_2 = \frac{V}{R_2} = \frac{40}{200} = 0.2 \ A$. Potential at $D$ relative to $A$ is $V_A - V_D = i_2 	imes 90 = 0.2 	imes 90 = 18 \ V$. Thus,$V_D = V_A - 18$. The potential difference across $BD$ is $|V_B - V_D| = |(V_A - 16) - (V_A - 18)| = |-16 + 18| = 2 \ V$.

Four resistances $40 \ \Omega, 60 \ \Omega, 90 \ \Omega$,and $110 \ \Omega$ form the arms of a quadrilateral $ABCD$. $A$ battery of emf $40 \ V$ with negligible internal resistance is connected across $AC$. Find the potential difference across $BD$. (in $V$)

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