as shown in figure a variable rheostat of 2 k ohm is used to control the potential difference across 500 ohm load. (1) if the resistance ABCA is 500 ohm
Answers
Answer ⇒ (a). 21.42 V and (b). 1600 ohm.
Explanation ⇒ Resistance along AB = 500 ohm.
Total Resistance of the rheostat = 2000 ohm.
Resistance of BC = 2000 - 500 = 1500 ohm.
Now, we can see resistance 1500 ohm and resistance of 500 ohm load is in parallel combination.
∴ Equivalent resistance of this parallel combination = 1500/4 = 375 ohm.
Now, First part of the rheostat AB and the Equivalent resistance of Parallel combination about BC are in series.
Thus, R equivalent = 500 + 375 = 875 ohm.
Now, current flowing in the circuit = Total Potential/Total resistance
= 50/875
= 2/35 A.
Now, In series current remains the same, so that current in part AB and in part BC is same.
Now, Potential given to part AB of Rheostat = IR = 2/35 × 500
= 1000/35 V.
Potential gone in part BC = 50 - 1000/35 = 21.42 V.
(b). When the load will be removed, R equivalent = 2000 ohm.
Thus, Total current flowing = 50/2000
= 5/200
= 2.5/100
= 0.025 A.
This same current will go in both part AB and BC.
Now, we want a potential of 40 V in part BC.
Thus, 40 = 0.025 × R
∴ 40 = 50/2000 × R
∴ R = 8 × 200
∴ R = 1600 Ohm.
Hope it helps.
Answer:
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