For the circuit shown in the diagram below:
what is the value of:
(i) current through 6 Ohm resistor? (ii) potential difference across 12 ohm resistor?
given:-V=4V, R1= 6 ohm, R2= 3 ohm, R3= 12ohm, R4= 3 ohm
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Answered by
727
(i) Applying Ohm's law,
let I be current 6 Ohm resistor
Then
I=4V/(6 Ohm +3 Ohm)=4/9 A
Current through both the resistors ohm and 3 ohm is 4/9 A as they are in series.
(ii) Let the current through 12 ohm and 3 ohm be I
then
I=4V/(12ohm+3ohm)=4/15A
Therefore potential difference across 12 ohm is given by
V=IR=4/15A x 12ohm=3.2V
let I be current 6 Ohm resistor
Then
I=4V/(6 Ohm +3 Ohm)=4/9 A
Current through both the resistors ohm and 3 ohm is 4/9 A as they are in series.
(ii) Let the current through 12 ohm and 3 ohm be I
then
I=4V/(12ohm+3ohm)=4/15A
Therefore potential difference across 12 ohm is given by
V=IR=4/15A x 12ohm=3.2V
Answered by
15
Answer:
As per the circuit
V = 4 V
Total resistance in line 1 = R1 = 6 + 3 = 9 ohm
Total resistance in line 2 = R2 = 12 + 3 = 15 ohm
(i) Current through 6 Ω resistor = Current through line 1 = V/R1 = 4/9 = 0.44 Ω
(ii) p.d. across line 2 is 4 V
Current through line 2 = V/R2 = (4/15) Ω
p.d. across 12 Ω resistor = (4/15) × 12 = 3.2 V
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