in the circuit given below,the potential difference across the arm PQ will be
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Please find below the solution to the asked query:
In the circuit the tow 6 ohm resistance are in parallel combination and the third 5 ohm resistance is in series combination with the above two. Hence the total resistance of the circuit is6×66+6+5=8 Ω.
So the current through the circuit is 248=3 A. So voltage drop across the parallel combination is 6×66+6×3=9 V. Since the voltage drop across PQ is equivalent to the voltage drop across the parallel combination, so the voltage drop across PQ is 9 V.
Hope this information will clear your doubts about electricity.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Please find below the solution to the asked query:
In the circuit the tow 6 ohm resistance are in parallel combination and the third 5 ohm resistance is in series combination with the above two. Hence the total resistance of the circuit is6×66+6+5=8 Ω.
So the current through the circuit is 248=3 A. So voltage drop across the parallel combination is 6×66+6×3=9 V. Since the voltage drop across PQ is equivalent to the voltage drop across the parallel combination, so the voltage drop across PQ is 9 V.
Hope this information will clear your doubts about electricity.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
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