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प्रश्न
Two lamps of resistance 30Ω and 20Ω respectively are connected in series in a 110V circuit. Calculate:
(i) the total resistance in the circuit
(ii) the current in the circuit, and
(iii) the voltage drop across each lamp.
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उत्तर
(i) Since the two lamps are connected in series, their total resistance R is given by:
R = R1 + R2
= 30 Ω + 20 Ω = 50 Ω
(ii) We have, V = 110 volt, R = 50 Ω
∴ By Ohm's law
I = `"V"/"R" = 110/50`A = 2.2 A
(iii) We now have
I = 2.2 A
∴ V1 = IR1 = 2.2 × 30V = 66 V
∴ Voltage frop across L1 = 66 V
Similarly, V2 = IR2 = 2.2 × 20 = 44 V
∴ Voltage drop across L2 = 44 V.
संबंधित प्रश्न
Two resistors of 4Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(i) Draw a labelled circuit diagram showing the above arrangement.
(ii) Calculate the current in each resistor. What is an Ohmic resistor?
Calculate the combined resistance in each case:
The diagram below shows part of a circuit:
If this arrangement of three resistors was to be replaced by a single resistor, its resistance should be:
(a) 9 Ω
(b) 4 Ω
(c) 6 Ω
(d) 18 Ω
Choose the correct alternative and rewrite the following sentence.
If three resistors 2 ohms, 3 ohms and 4 ohms are connected in series, then the effective resistance in the circuit will be _______ ohms.
Write an expression for calculating electrical power in terms of current and resistance.
Two resistors having 2Ω and 3Ω resistance are connected—(i) in series, and (ii) in parallel. Find the equivalent resistance in each case.
What is the maximum resistance which can be made using five resistors each of `1/5` W?
If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be:
