# Electric circuit question [closed]

You have to calculate Id when the switch at S is opened, when its closed, the Id = 0.5 A. All resistors are the identical in the circuit.

So, the answer is B, Id = 0.67 A. But why? I don't get it, maybe its elementary, but I don't get it. Could you explain to me, in steps, how to solve this problem?

Thanks.

-

## closed as off-topic by Emilio Pisanty, akhmeteli, dmckee♦Sep 22 '13 at 16:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, akhmeteli, dmckee
If this question can be reworded to fit the rules in the help center, please edit the question.

Please note that Homework questions are supposed to show more effort. For more details, please see the Homework Policy. – centralcharge Sep 22 '13 at 10:19

Start with the switch closed. You're told that all the resistors have the same value - let's call this $R$. You can use the usual rules for calculating the total resistance to calculate $R_{total}$. Now the battery voltage is $V = IR_{total}$ and you know $I$ so you can calculate $V$ in terms of $R$.
With the switch open you calculate $R_{total}$ again (obviously it's higher than with the switch closed), and you calculated $V$ above, so now you can calculate $I$.