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 $E$?$\huge$ $\endgroup$ – Brian Moths Nov 25 '14 at 15:19
  • $\begingroup$ The electromotive force of the battery:) $\endgroup$ – user9686 Nov 25 '14 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. $\endgroup$ – Brian Moths Nov 25 '14 at 15:47

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.

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  • $\begingroup$ 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 '14 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? $\endgroup$ – Brian Moths Nov 25 '14 at 15:35

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