The terminals of a battery of emf 24.0 V and internal resistance of 1.0 Ω is connected to an external resistor 5.0 Ω. Find the terminal p.d.
- 18.0V
- 12.0V
- 16.0V
-
20.0V
The correct answer is: D
Explanation
ε = 24.0 V, r=1Ω, R=5.0Ω
terminal p·d = ?
ε = IR+Ir = I(R+r)
⇒ IR+Ir = I(R+r)
⇒ I = \(\frac{24}{5+1}\)
⇒ I =\(\frac{24}{6}\)
⇒ I = 4 A
terminal p·d = 4 × 5
terminal p·d = 20.0 V
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