A6.0 v battery delivers a current of 120 ma to a 46 ω resistor. the voltage at the battery terminals is: a.4.2 v
b.5.5 v
c.9.6 v
d.8.4 v

V= 5.52 V

Explanation:

Given that

V= 6 V

I =  120 mA

R= 46 Ω

We know that

ΔV= R I

ΔV= 120 x 0.001 x 46

V= 5.52 V

So the option b is correct.

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