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Correct diagram for the determination of internal resistance of a primary cell by potentiometer is

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Important Questions on Current Electricity

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A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance r. A cell C having EMF ε/2 and internal resistance 3r is connected. The length AJ, at which the galvanometer, as shown in the figure, shows no deflection is

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AB is a wire of potentiometer with the increase in the value of resistance R, the shift in the balance point J will be
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In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120 cm. The  internal resistance of the cell is
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A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of EMF's is :
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A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves:
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A potentiometer has a uniform wire of length 5 m. A battery of emf 10 V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200 cm. The emf of the secondary cell is
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The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA For a cell of emf 5 V and internal resistance of 20 Ω the null point on it is found to be at 1000 cm The resistance of whole wire is:
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In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells ε1ε2 is:

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In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω ,a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.
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In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B . The resistance per unit length of the potentiometer wire is r=0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J  at 50 cm from end A , the expected reading of the voltmeter will be:
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If the balancing length in a potentiometer experiment changes by 10 % when the cell of unknown emf is shunted by 6 Ω resistance, then the internal resistance of the cell in Ω is
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A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E0 and a resistance r1 . An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by:
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In a potentiometer, a wire of length 10 m having resistance 50 Ω is used. A battery of 5 V and a resistor of 450 Ω are connected in series to the wire. If an unknown battery of emf 'E' balances the potentiometer at 450 cm, then the value of E is
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A potentiometer wire PQ of 1 m length is connected to a standard cell E1. Another cell E2 of emf 1.02 V is connected with a resistance r and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is :

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Figure shows a potentiometer. Length of the potentiometer wire AB is 100 cm and its resistance is 100Ω. EMF of the battery E is 2 V. A resistance R of 50Ω draws current from the potentiometer. What is the voltage across R when the sliding contact Cis at the mid-point of AB ?

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In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
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In the given figure, there is a circuit of potentiometer of length AB=10 m. The resistance per unit length is 0.1 Ω per cm. Across AB, a battery of emf E and internal resistance r is connected. The maximum value of emf measured by this potentiometer is:

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A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of.
i infinity ii 9.5 Ω
The 'balancing lengths, on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of internal resistance of the cell is:
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A potentiometer PQ is set up to compare two resistances, as shown in the figure. The ammeter A in the circuit reads 1.0 A when the two-way key K3 is open. The balance point is at a length l1 cm from P when the two-way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when the key K3 is plugged in between 3 and 1 . The ratio of two resistances R1R2, is found to be 
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A potentiometer wire has length 4 m and resistance 8Ω . The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V , so as to get a potential gradient 1 mV per cm on the wire is :