Advertisements
Advertisements
Questions
Answer the following question.
A cell of emf E and internal resistance r is connected across a variable resistor R. Plot the shape of graphs showing a variation of terminal voltage V with (i) R and (ii) circuit current I.
A cell emf of (E) and internal resistance. (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage V with (i) R and (ii) the current (I) in the load.
Advertisements
Solution
Terminal Voltage V can be related to R by the following relation:
(i) `V = (ER)/(R+r)`
⇒ `V = (E)/(1+r/R)`

(ii) Terminal Voltage V can be related to I by the following relation:
V = E - IR

APPEARS IN
RELATED QUESTIONS
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.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
Two cells of emf E1, E2 and internal resistance r1 and r2 respectively are connected in parallel as shown in the figure.

Deduce the expressions for
(1) the equivalent e.m.f of the combination
(2) the equivalent resistance of the combination, and
(3) the potential difference between the point A and B.
Can the potential difference across a battery be greater than its emf?
A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits.
Find its value. This is the basic principle of a constant-current source.
A coil of resistance 100 Ω is connected across a battery of emf 6.0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0 J K−1, how long will it take to raise the temperature of the coil by 15°C?
Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
Five cells each of emf E and internal resistance r send the same amount of current through an external resistance R whether the cells are connected in parallel or in series. Then the ratio `("R"/"r")` is:
An ac generator generates an emf which is given by e = 311 sin (240 πt) V. Calculate:
- frequency of the emf.
- r.m.s. value of the emf.
