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
waec 2022physics

The correct answer is C. 0.30A

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 Ã· 3.67 = I

0.27 = I

: I â‰ˆ 0.30A

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