There are two electric circuits.

  • Resistor 1 ($R_1$) connected with resistor 2 ($R_2$) by series circuit
  • $R_1$ connected with $R_2$ in parallel

Each of the circuit has battery with emf and the internal resistance $r$. The ammeter is connected to the circuit before the junction. The voltmeter is connected to the circuit by bestride both resistances.

The ammeter in circuit one reads $I_1$, voltmeter reads $V_1$. The ammeter in circuit 2 reads $I_2$, voltmeter reads $V_2$. The question ask to compare between $V_1$ and $V_2$, $I_1$ and $I_2$.

I think the answer should be $V_1$ is lesser than $V_2$ and $I_1$ lesser than $I_2$, because $\Delta V= Ir+E$ ,so $V$ depends on $I$. But the solution is $V_1$ more than $V_2$. Do I understand something wrong?

  • 1
    $\begingroup$ what is $E$?$\huge$ $\endgroup$ Nov 25, 2014 at 15:19
  • $\begingroup$ The electromotive force of the battery:) $\endgroup$
    – user9686
    Nov 25, 2014 at 15:46
  • $\begingroup$ Then $\Delta V = E-Ir$ because voltage is lost across the internal resistor. That is why the internal resistance is a bad thing. $\endgroup$ Nov 25, 2014 at 15:47

1 Answer 1


I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its $V$ (notice that $V$ is really just measuring the voltage drop across the battery) has to be lower.

  • $\begingroup$ I don't understand why more stressed, and it gives out a lower voltage. From V=IR, more current, more voltage(v is direct variation to I)? $\endgroup$
    – user9686
    Nov 25, 2014 at 15:31
  • $\begingroup$ It's more stressed because it has to output more current. I thought you said the answer said the parallel $V$ was less than the series one? $\endgroup$ Nov 25, 2014 at 15:35

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