As shown in the figure initially switch $k_{1}$ is open and $k_{2}$ is closed and deflection of galvanometer is $θ_{0}$ . Now both switch are closed then the deflection in galvanometer is (given $R_{H} = 5 Ω$ )

Question

As shown in the figure initially switch k_1 is open and k_2 is closed and deflection of galvanometer is θ_0 . Now both switch are closed then the deflection in galvanometer is (given R_H=5 Ω )

MARKS Content Team · Accepted Answer

Correct answer: (B) 22 Ω. θ_0∝i_G⇒Cθ_0=i_G I-case Cθ_0=v/(220+R) ......(1) II-case C(θ_0)/5v/(220+5R5+R)×5/(5+R) ..........(2) From (1) and (2) R=22 Ω

As shown in the figure initially switch $k_{1}$ is open and $k_{2}$ is closed and deflection of galvanometer is $θ_{0}$ . Now both switch are closed then the deflection in galvanometer is (given $R_{H} = 5 Ω$ )

Solution

More Experimental Physics Questions

As shown in the figure initially switch k1 is open and k2 is closed and deflection of galvanometer is θ0 . Now both switch are closed then the deflection in galvanometer is (given RH=5 Ω )

Solution

More Experimental Physics Questions

As shown in the figure initially switch $k_{1}$ is open and $k_{2}$ is closed and deflection of galvanometer is $θ_{0}$ . Now both switch are closed then the deflection in galvanometer is (given $R_{H} = 5 Ω$ )