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In the given circuit, i need to find the current flowing in the circuit when the switch is closed and for this i need to calculate the equivalent resistance between the terminals of the cell. How would you calculate the equivalent resistance between the terminals of the cell?

I calculated the equivalent resistance to be by taking 6 ohms and 3 ohms resistors in parallel and the equivalent resistance of these two in series with the 10 ohms resistor. The equivalent resistance between the terminals of the cell comes out to be 2 ohms and the current hence comes out to be 5/6 A which is not present in the options. (The options are-:2 A, 1 A, 3 A and 0 A.)

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    \$\begingroup\$ There is something wrong with your series calculation. Also, think carefully about what the switch does. \$\endgroup\$ Commented Dec 27, 2016 at 21:35
  • \$\begingroup\$ @PeterBennett Oh i meant 5/6 ohms. \$\endgroup\$
    – MrAP
    Commented Dec 27, 2016 at 21:38
  • \$\begingroup\$ Current is not measured in Ohms. You still seem to have a problem with claculating resistors in series. \$\endgroup\$ Commented Dec 27, 2016 at 21:41
  • \$\begingroup\$ @PeterBennett I meant that the equivalent resistance is 12 ohms and the current is 5/6 A. Sorry for that. \$\endgroup\$
    – MrAP
    Commented Dec 27, 2016 at 21:46
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    \$\begingroup\$ With the switch open, there is no current through the switch, so it has no effect on the circuit. In that case, you have the 6 and 3 ohm resistors in parallel, for an equivalent resistance of 2 Ohms. That is in series with the 10 Ohm, for a total circuit resistance of 12 Ohms. \$\endgroup\$ Commented Dec 27, 2016 at 23:17

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When the switch is shorted there's no current through the 6 Ohm and the 3 Ohm resistors so the current value would be 10/10=1A.

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    \$\begingroup\$ On this sort of question, I try to give clues that will help the poster find answer, rather than just giving him the answer. \$\endgroup\$ Commented Dec 27, 2016 at 22:45
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    \$\begingroup\$ Exactly. The question asked 'how do you find ...?', not 'what is...?' \$\endgroup\$
    – Chu
    Commented Dec 28, 2016 at 0:48

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