Question

As shown in figure a variable rheostat of 2 kis used to control the potential difference across 500 ohm load. (i) If the resistance AB is 500 , what is the potential difference across the load? (ii) If the load is removed, what should be the resistance at BC to get 40 volt between B and C

Answer 1

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