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Three resistors of 10 Omega, 15 Omegaand 20 Omega are connected in series in a circuit. If the potential drop across the 15 Omega resistor is 3 V, find the current in the circuit and potential drop across the 10 Omega resistor. |
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Answer» SOLUTION :In series circuit same current FLOWS through all the resistors. Current through 15 `Omega` resistor `I = (V)/(R) = (3V)/(15) = (1)/(5)` =0.2 A `:.` Current in the circuit `=0.2` A `:.` Potential drop across 10`Omega` resistor is `I =(V)/(R)` `V= IR` `=0.2A xx10 Omega` =2 V |
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