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Compare the power used in the 2Omega resistor in each of the following circuits: i] a 6 V battery in series with 1Omegaand2Omega resistors, and ii] a 4 V battery in parallel with 12Omegaand2Omega resistors. |
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Answer» Solution :i] Potential difference M = 6 V Here `1Omegaand2Omega` resistors are connected in series, `therefore` equivalent resistance of the circuit [R] = 1 + 2 = `3Omega`, According to Ohm.s law `V=IRorI=V/R=6/3=2A` the CURRENT through the circuit [I] = 2A Now the power USED in the `2Omega` resistor Power = `I^(2)R=2xx2xx2=8W` ii] Potential difference, V = 4V `12Omegaand2Omega` resistors are connected in PARALLEL. The voltage across each component of a parallel circuit REMAINS the same. Hence, the voltage across `2Omega` resistor will be 4 V. Power consumed by `2Omega` resistor is given by `P=V^(2)/R=4^(2)/2=8W` `therefore` the power used by `2Omega` resistor is 8W. |
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