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A battery of emf E has an internal resistance r. A variable resistance R is connected to the terminals of the battery. A current `i` is drawn from the battery. V is the terminal potential difference. If R alone is gradually reduced to zero, which of the following best describes `i and
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A cell of emf (E) and internal resistance r is connected across a variable external resistance R . The graph of terminal potential difference V as a function of R
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In the electric circuit shown, each cell has an emf of `2 V` and internal resistance is `1Omega` . The external resistance is `2Omega`. The value of the current `I` is (in
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Consider first two cells in series as shown in figure. The potential difference between the terminals A and C of the combination
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In the circuit shown in figure, `E_1=10V, E_2=4 V , r_1 = r_2 = 1Omega and R = 2Omega` Find the potential difference across battery 1 and battery
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Under what conditions current passing through the resistance R can be increased by short circuiting the battery of emf `E_2` ? The internal resistances of the two batteries are `r_1 and r_2`,
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12 cells, each of emf `1.5 V` and internal resistance of `0.5 Omega`, are arranged in `m` rows each containing `n` cells connected in series, as shown in the figure.Calculate the values of `n and m` for which this combination would send maximum current through an external resistance of `1.5
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According to Kirchhoff’s current law as applied to a junction in a network of
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Five conductors are meeting at a point `x` as shown in the figure. What is the value of current in fifth
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The currents `i_1 and i_2` through the resistors `R_1(=10 Omega) and R_2 (= 30 Omega)` in the circuit diagram with `E_1= 3 V, E_2 = 3 V and E_3 = 2 V` are
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The algebraic sum of voltages around any closed path in a network is equal
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The potential difference `(V_A - V_B)` between the points A and B in the given figure
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If `2A` current is flowing in the shown circuit, then potential difference `(V_B - V_D)` in balanced condition
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In a Wheatstone bridge circuit, `P=7 Omega, Q= 8 Omega, R = 12 Omega and S = 7 Omega`. Find the additional resistance to be used in series with `S` , so that the bridge is
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A meter bridge is set up as shown in figure, to determine an unknown resistance `X` using a standard `10 Omega` resistor. The galvanometer shows null point when tapping key is at `52 cm` mark. The end-corrections are `1 cm and 2 cm` respectively for the ends A and B. The determined value of `X`
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The figure below shows a `2.0 V` potentiometer used for the determination of internal resistance of a `2.5 V` cell. The balance point of the cell in the open circuit is 75 cm. When a resistor of `10 Omega` is used in the external circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. The internal resistance of the cell
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A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire, is `k` volt/cm and the ammeter, present in the circuit, reads `1.0` A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths `l_1 cm and l_2 cm`, respectively. The magnitudes of the resistors `R and X` in ohm, are then, respectively equal
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`10 A` current passes through a copper. The net charge after `5s` in the wire
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Au is a highly conductive metal which obeys Ohm’s law but................does not obey Ohm’s law due to lack of conduction
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During charging of a cell terminal voltage i s ................ than emf of
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