The internal resistance of the cell shown in the figure is negligible. On closing the key $k$ , the ammeter reading changes from $0.25 A$ to $\frac{5}{12} A$ , then what is the value of resistance $R_{1}$ (in ohm)?

Question

The internal resistance of the cell shown in the figure is negligible. On closing the key k, the ammeter reading changes from 0.25 A to 5/12 A, then what is the value of resistance R_1 (in ohm)?

MARKS Content Team · Accepted Answer

Correct answer: 15. When k was open Reading of A =E/10 A = 0.25, When k was closed Reading of A =(E10+R_1)/10R_1=5/12 ⇒ 12 5 + 0.5R_1 =10R_1 R_1 = 15 Ω

The internal resistance of the cell shown in the figure is negligible. On closing the key $k$ , the ammeter reading changes from $0.25 A$ to $\frac{5}{12} A$ , then what is the value of resistance $R_{1}$ (in ohm)?

Solution

More Current Electricity Questions

The internal resistance of the cell shown in the figure is negligible. On closing the key k, the ammeter reading changes from 0.25 A to 512 A, then what is the value of resistance R1 (in ohm)?

Solution

More Current Electricity Questions

The internal resistance of the cell shown in the figure is negligible. On closing the key $k$ , the ammeter reading changes from $0.25 A$ to $\frac{5}{12} A$ , then what is the value of resistance $R_{1}$ (in ohm)?