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(B) Let the total current from the battery be $i$ and the current through the ammeter be $i_1$. The current through the $4\, \Omega$ resistor is $(i -

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$40/29\, A$

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(B) Let the total current from the battery be $i$ and the current through the ammeter be $i_1$. The current through the $4\, \Omega$ resistor is $(i - i_1)$.Applying Kirchhoff's voltage law to the outer loop:$10 - 1i - 4(i - i_1) = 0$$10 = 5i - 4i_1$...... $(i)$Applying Kirchhoff's voltage law to the loop containing the ammeter and the $4\, \Omega$ resistor:$5i_1 - 4(i - i_1) = 0$$5i_1 = 4i - 4i_1$$9i_1 = 4i \implies i = \frac{9}{4}i_1$...... $(ii)$Substituting $(ii)$ into $(i)$:$10 = 5(\frac{9}{4}i_1) - 4i_1$$10 = \frac{45}{4}i_1 - \frac{16}{4}i_1$$10 = \frac{29}{4}i_1$$i_1 = \frac{40}{29}\, A$

If the internal resistance of the battery is $1\, \Omega$, then what is the reading of the ammeter?

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