For the circuit shown in the figure,
I=0.50 A and R= 12 N. What is the
value of the emf e?
R
2R
2R
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Answer:
(i) R=6+3=9Ω
V=4V
I=
R
V
=
9
4
=0.45A
(ii) 12 ohm and 3 ohm are in series
R=12+3=15Ω
I=
R
V
=
15
4
=0.27A
p.d across 12 ohms= IR=0.27×12=3.24V
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