InterviewSolution
Saved Bookmarks
| 1. |
Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Omegaresistor, a 10 Omega resistor and a 15 Omega resistor, an ammeter and a plug key, all connected in aeries. Also connect a voltmeter to record tht: potential difference across the 15 Omega resistor and calculate : (i) the electric current passing through the above circuit and (ii) potential difference across 5 Omega resistor when the key is closed. |
Answer» SOLUTION :(i) 10 V battery Rest COMPONENTS Equivalent resistance =`R_(1)+R_(2)+R_(3)` `=5+10+15` `=30 Omega` Current in the CIRCUIT `I= (V)/(R)` `I = (10V)/(30Omega) = (1)/(3) A ` or 0.33 A (ii) Potential difference across 5 `Omega` resistor V= IR `=(1)/(3)A xx5 Omega` 1.67 V |
|