Advertisements
Advertisements
प्रश्न
Explain, why is the p.d. between the terminals of a storage battery less when it is supplying current than when it is on open circuit. A battery of e.m.f. 10 volts and internal resistance 2.5 ohms has two resistances of 50 ohms each connected to it. Calculate the power dissipated in each resistance
(a) When they are in series,
(b) When they are in parallel.
In each case calculate the power dissipated in the battery.
Advertisements
उत्तर
In an open circuit no current is drawn from the cell whereas in closed circuit, an amount of energy is spent in the flow of unit positive charge through the electrolyte of the cell. Thus, the p.d. between the terminals of a storage battery is less when it is supplying current.
Given, emf e = 10 V, internal resistance r = 2.5 Ω
Two external resistances R1 = R2 = 50Ω (= R, say)
(a) When R1 and R2 are connected in series, same current flows through each resisto and hence same power is dissipated in each resistor.
Total resistance of the circuit, Rs=R1 + R2 +r = 50 + 50 + 2.5 = 102.5Ω
Current through each resistor = `"e"/"R"_"s" = 10/102.5 = 0.096"A"`
Power dissipated in each resistor =I2R (0.096)2 (50) = 0.46 watt
(b) When R1 and R2 are connected in parallel, voltage across each resistor is same and hence same power is dissipated in each resistor.
Total resistance of the circuit Rp = `(1/"R"_1 + 1/"R"_2)^-1 + r = 25 + 2.5 = 27.5 Omega`
Current through the circuit, I= `"e"/"R"_"p" = 10/27.5 = 0.36 "A"`
Voltage across each resistor, V = e -Ir= 10 - ( 0.36 x 2.5) = 9.1V
Power dissipated in each resistor P = `"V"^2/"R" = (9.1)^2/50 = 4140.5 "watt"`
APPEARS IN
संबंधित प्रश्न
With a neat labelled diagram and derive the equation for three resistances connected in parallel.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
A wire that has resistance R is cut into two equal pieces. The two parts are joined in parallel. What is the resistance of the combination?
How will you connect three resistors of resistances 2 Ω, 3 Ω, and 6 Ω to obtain a total resistance of 4 Ω, and 1 Ω?
State how are the two resistors joined with a battery when equivalent resistance is less than either of the two resistances.
You have three resistors of values 2Ω, 3Ω and 5Ω. How will you join them so that the total resistance is less than 1Ω? Draw diagram and find the total resistance.
The circuit diagram Fig shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f. 2 V and internal resistance 3 Ω. If main current of 0.25 A flows through the circuit, find:
- the p.d. across the 4 Ω resistor,
- the p.d. across the internal resistance of the cell,
- the p.d. across the R Ω or 2 Ω resistors
- the value of R.
Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
Draw a labeled diagram of the arrangement
In the figure below, the ammeter A reads 0.3 A. Calculate:
(i) the total resistance of the circuit
(ii) the value of R
(iii) the current flowing through R.
Calculate equivalent resistance in the following cases:
