Advertisements
Advertisements
Question
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.
Advertisements
Solution
Let I1 and I2 be the current in two cells with emf E1 and E2 and internal resistance r1 and r2
So I = I1 + I2
Now let V be the potential difference between the points A and B. Since the first cell is connected between the points A and B.
V = potential difference across first cell
V = E1 − I1r1
`or I_1 =(E_1-V)/r_1`
Now, the second cell is also connected between the points A and B. So,
`or I_2 =(E_2-V)/r_2`
Thus, substituting for I1 and I2
`I =(E_1-V)/r_1 + (E_2-V)/r_2`
`or, I (E_1 /r_1 + E_2 /r_2)-V(1 /r_1 +1 /r_2)`
`V =((E_1r_2 + E_2r_1)/(r_1+r_2)) -I((r_1r_2)/(r_1 +r_2)) ..... (1)`
If E is effective e.m.f and r, the effective internal resistance of the parallel combination of the two cells then V = E − Ir …..(2)
Comparing (1) and (2)
`(1) E= (E_1r_2 + E_2r_1)/(r_1+r_2)`
This is equivalent e.m.f of the combination
`(2)r =(r_1r_2)/(r_1 +r_2)`
This is equivalent resistance of the combination.
(3) The potential difference between the point A and B is
V = E − Ir
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?
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.)
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.
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
A cell of emf E and internal resistance r is connected to two external resistance R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
Apply the first law of thermodynamics to a resistor carrying a current i. Identify which of the quantities ∆Q, ∆U and ∆W are zero, positive and negative.
If n cells each of emf e and internal resistance r are connected in parallel, then the total emf and internal resistance will be ______.
A block of metal is heated directly by dissipating power in the internal resistance of block. Because of temperature rise, the resistance increases exponentially with time and is given by R(t) = 0.5 e2t, where t is in second. The block is connected across a 110 V source and dissipates 7644 J heat energy over a certain period of time. This period of time is ______ × 10-1 sec (take ln 0.367 = -1).
