In the figure each cell has an emf of 1.5 V and internal resistance of 0.40 Ω Ω . Calculate:

(i) current I
(ii) current in the 36 Ω Ω resistor
(iii) potential difference across A and B.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
[(i)
= 0.5 A
(ii)
= 0.0833 A
(iii)
= 1.7 V]
Sol. (i) Effective emf = 4 × 1.5 – 1.5 = 4.5 V
(ii) Effective internal resistance
r eq =
+ 0.4 = 1.2 Ω Ω
R eq = 6 + 1.8 + 1.2 = 9 Ω Ω
I =
amp
I 36 = 
(iii) V AB = E + ir = 1.5 +
× 0.4 = 1.5 + 0.2 = 1.7 V
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Six lead-acid type of secondary cells, each of emf 2.0 V and internal resistance 0.015 Ω Ω , are jo…
In the circuit shown all five resistors have the same value 200 ohms and each cell has an emf 3 vol…
Find the currents through the three resistors shown in figure
12 cells each having the same emf are connected in series and are kept in a closed box. Some of the…
Two cells of e.m.f. 10 V & 15 V are connected in parallel to each other between points A & B. The c…
N sources of current with different emf ’ s are connected as shown in figure. The emf ’ s of the so…