Three cells each of emf, 1.0 V, and internal resistance, 2 \(\Omega\), are connected in parallel across a 3\(\Omega\) resistor. Determine the current in the resistor.
A
0.90A
B
0.39A
C
0.30A
D
0.01A
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?
3 resistance in parallel = \(\frac{1}{r_1}\) + \(\frac{1}{r_2}\) + \(\frac{1}{r_3}\)
\(\frac{1}{r_T}\)= \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
\(\frac{1}{r_T}\) = \(\frac{1+1+1}{2}\)
\(\frac{1}{r_T}\) = \(\frac{3}{2}\)
cross multiply
r\(_T\) = \(\frac{2}{3}\) or 0.67
E = I(R+r)
1 = I(3+0.67)
1 = I(3.67)
1 \div 3.67 = I
0.27 = I
: I ≈ 0.30A
Source: WAEC / JAMB / NECO Classroom Archive (Record #63324)