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Solution - Two Cells of Emfs E1 and E2 and Internal Resistances R1 and R2 Are Connected in Parallel. Derive the Expression for the (I) Emf and (Ii) Internal Resistance of a Single Equivalent Cell Which Can Replace this Combination. - Induced Emf and Current

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Two cells of emf E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive the expression for the (i) emf and (ii) internal resistance of a single equivalent cell which can replace this combination.

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A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−1 along the negative x-direction (that is it increases by 10− 3 T cm−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3 T s−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

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Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

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Solution for question: Two Cells of Emfs E1 and E2 and Internal Resistances R1 and R2 Are Connected in Parallel. Derive the Expression for the (I) Emf and (Ii) Internal Resistance of a Single Equivalent Cell Which Can Replace this Combination. concept: Induced Emf and Current. For the courses 12th CBSE (Arts), 12th CBSE (Commerce), 12th CBSE (Science)
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