In the circuit shown,the cell has emf = 10,V | vedclass

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(D) To find the current,we simplify the circuit from right to left.$1$. The rightmost loop consists of a $2\, \Omega$ resistor in series with a $4\, \

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The current through the $4\, \Omega$ resistor is $0.25\,A$

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(D) To find the current,we simplify the circuit from right to left.$1$. The rightmost loop consists of a $2\, \Omega$ resistor in series with a $4\, \Omega$ resistor,which is in series with another $2\, \Omega$ resistor. However,looking at the circuit,the $4\, \Omega$ resistor is in series with the $2\, \Omega$ and $2\, \Omega$ resistors in the rightmost branch,giving $R = 2+4+2 = 8\, \Omega$. This $8\, \Omega$ is in parallel with the vertical $8\, \Omega$ resistor,resulting in $R_p = 4\, \Omega$.$2$. Now,this $4\, \Omega$ is in series with the horizontal $2\, \Omega$ and $2\, \Omega$ resistors,giving $R = 2+4+2 = 8\, \Omega$. This $8\, \Omega$ is in parallel with the middle vertical $8\, \Omega$ resistor,resulting in $R_p = 4\, \Omega$.$3$. Finally,this $4\, \Omega$ is in series with the horizontal $3\, \Omega$ and $2\, \Omega$ resistors and the internal resistance $1\, \Omega$,giving $R_{total} = 3+4+2+1 = 10\, \Omega$.$4$. The total current from the battery is $I = \frac{10\,V}{10\, \Omega} = 1\,A$.$5$. By current division,the current splits equally at each junction where the resistances are equal. The current through the $4\, \Omega$ resistor is $0.25\,A$.

In the circuit shown,the cell has $emf = 10\,V$ and internal resistance $= 1\, \Omega$. Find the current through the $4\, \Omega$ resistor.

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