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प्रश्न
Two cells each having an e.m.f. of 2 V and an internal resistance of 2Ω are
connected (a) In series, and ( b) In para 1le l as shown in fig. . What is the
current flowing through the cir cu it in each_ case?

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उत्तर
(a) Given, emf (e) of the battery= 2 x 2 = 4V,
internal resistance r = 2 x 2= 4Ω
Resistance in the circuit, R = 4 Ω
Total resistance of the given series circuit = 4 + 4 = 8Ω
Current flowing through the circuit, I = `"e"/("R" + "r") = 4/8 =0.5 "A"`
(b) Given, emf of battery ( e) = emf of each cell in parallel = 2V,
Total internal resistance `1/"r" = 1/2 + 1/2 = 1Ω`
Resistance connected in the circuit ,R = 4Ω
Total resistance of the circuit= 1 + 4 = 5 Ω
Current flowing through the circuit, I= `"e"/("R" + "r") = 2/5 = 0.4 "A"`
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संबंधित प्रश्न
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(a) current
(b) potential difference
(c) resistance
(d) power
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If the current through a floodlamp is 5 A, what charge passed in 10 seconds?
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In the network shown in the following adjacent Figure, calculate the equivalent resistance between the points.
- A and B
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Two resistors of 2.0 Ω and 3.0 Ω are connected (a) in series (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Conventionally, the direction of the current is taken as:
Match the items in column-I to the items in column-II:
| Column - I | Column - II |
| (i) electric current | (a) volt |
| (ii) potential difference | (b) ohm meter |
| (iii) specific resistance | (c) watt |
| (iv) Electrical power | (d) joule |
| (v) electrical energy | (e) ampere |
