WAEC - Physics (2022 - No. 23)

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.
0.90A
0.39A
0.30A
0.01A

Explanation

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 ÷ 3.67 = I

0.27 = I

: I ≈ 0.30A

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