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A battery of emf `E_(0) = 12 V` is connected across a `4 m` long uniform wire having resistance `(4 Omega)/(m)`. The cells of small emfs `epsilon_(1) = 2 V` and `epsilon_(2) - 4 V` having internal resistance `2 Omega` and `6 Omega` respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point `N`, the distance of point `N` from teh point `A` is equal to A. `1//6m`B. `1//3 m `C. ` 25 cm `D. ` 50 cm ` |
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Answer» Correct Answer - C c. `R_(AB) = 16 Omega, V_(AB) = (R_(AB)epsilon_0)/(R+R_(AB)) = (16 xx 12)/(8+16) = 8V`. Current in the loop of `epsilon_1 and epsilon_2` : `I = (epsilon_2 + epsilon_1)/(r_1+r_2) = (4+2)/(2+6) = 3/4 A` `epsilon_1 - Ir_1 = V_(AB) = (AN)/(AB) (V_(AB)) or 2-3/4 xx 2 = (AN)/4 xx 8 ` or `AN = 1/4 m = 25 cm`. |
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