Find the potential difference between points | vedclass

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(B) The circuit consists of two cells of $EMF$ $6 \ V$ and $4 \ V$ connected in opposition,and two resistors of $2 \ \Omega$ and $8 \ \Omega$ in serie

Vedclass · Accepted Answer

$5.6$

Vedclass · Answer

(B) The circuit consists of two cells of $EMF$ $6 \ V$ and $4 \ V$ connected in opposition,and two resistors of $2 \ \Omega$ and $8 \ \Omega$ in series.Net $EMF$,$E_{net} = 6 \ V - 4 \ V = 2 \ V$.Total resistance,$R_{total} = 2 \ \Omega + 8 \ \Omega = 10 \ \Omega$.The current in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{2 \ V}{10 \ \Omega} = 0.2 \ A$.To find the potential difference between $A$ and $B$,we move from $A$ to $B$ through the $4 \ V$ cell and the $8 \ \Omega$ resistor.The potential difference across the $8 \ \Omega$ resistor is $V_R = I 	imes R = 0.2 \ A 	imes 8 \ \Omega = 1.6 \ V$.Moving from $A$ to $B$,we encounter the $4 \ V$ battery (from positive to negative terminal,so potential drops) and the resistor (potential drops in the direction of current).$V_A - V_B = 4 \ V + (I 	imes 8 \ \Omega) = 4 \ V + 1.6 \ V = 5.6 \ V$.

Find the potential difference between points $A$ and $B$ in the following circuit. (in $V$)

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