Advertisements
Advertisements
Question
A cell of emf 'E' and internal resistance 'r' is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I.
It is found that when R = 4 Ω, the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.
Advertisements
Solution
(i)
Plot for V vs. R
(ii)
When R = 4 Ω and I = 1 A
We know
Terminal voltage, V = E − Ir
So, we have
V = IR = 4 = E − Ir
or
E − r = 4 ..... (i)
When R' = 9 Ω and I' = 0.5 A
V = IR = 0.5 × 9 = E − 0.5r
or
E − 0.5r = 4.5 ..... (ii)
Solving (i) and (ii), we get
r = 1 Ω and E = 5 V
APPEARS IN
RELATED QUESTIONS
A battery of emf 12 V and internal resistance 2 Ω is connected to a 4 Ω resistor as shown in the figure.
(a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.
(b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?
Distinguish between emf and terminal voltage of a cell.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
A resistor R is connected to a cell of-emf e and internal resistance r. The potential difference across the resistor R is found to be V. State the relation between e, V, Rand r.
Find the value of i1/i2 in the following figure if (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.
The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0°C and 0.001°C. Find the thermo-emf (Seebeck emf) developed. For bismuth-silver, a = − 46 × 10−6 V°C−1 and b = −0.48 × 10−6 V°C−2.
A plate of area 10 cm2 is to be electroplated with copper (density 9000 kg m−3) to a thickness of 10 micrometres on both sides, using a cell of 12 V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100 g of water, calculate the rise in the temperature of the water. ECE of copper = 3 × 10−7 kg C−1and specific heat capacity of water = 4200 J kg−1.
Internal resistance of a cell increases when the distance between electrodes:
