A battery of emf 15 V and internal resistence 3 ohm is connected to two resistors of resistance 3 ohm and 6 ohm in parallel . Find the potential difference between the terminal of the battery
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Here Emf (V) = 15V
internal resistance (r) = 3ohm
R1= 3ohm
R2= 6ohm
Now
Total resistance = r+ (R1×R2)/(R1+R2) = 3+(3×6)/(3+6) = 3+18/9 = 3+2 = 5ohm
Now current (i) = V/R = 15/5 = 3A
p.d between terminals = emf-i×internal resistance = 15-3×3 = 15-9 = 6V is right answer. hope it helps you
internal resistance (r) = 3ohm
R1= 3ohm
R2= 6ohm
Now
Total resistance = r+ (R1×R2)/(R1+R2) = 3+(3×6)/(3+6) = 3+18/9 = 3+2 = 5ohm
Now current (i) = V/R = 15/5 = 3A
p.d between terminals = emf-i×internal resistance = 15-3×3 = 15-9 = 6V is right answer. hope it helps you
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