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Homework answers / question archive / Part A For the circuit shown in the figure(Figure 1) find the current through each resistor
Part A
For the circuit shown in the figure(Figure 1) find the current through each resistor.
Express your answers using two significant figures. Enter your answers numerically separated by commas.
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I3Ω, I24Ω, I5Ω, I4Ω, I12Ω = | A |
Part B
For the circuit shown in the figure find the potential difference across each resistor.
Express your answers using two significant figures. Enter your answers numerically separated by commas.
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ΔV3Ω, ΔV24Ω, ΔV5Ω, ΔV4Ω, ΔV12Ω = | V |
24 2 4 5 n WW 12 V 12 (2 3 V WW 3 n
here 4 ohm and 12 ohm resistance are in parellel
so the equilent resistance
= (4x12) / (4+12)
= 3 ohm
now 3 ohm and 5 ohm are in series
so 3 + 5 = 8 ohm
now 8 ohm and 24 ohm are in parallel
so the equivalent resistance = (8x24) / (8+24)
= 6 ohm
now apply KVL in loop
12 - 6I - 3 - 3I = 0
9I = 9
I = 1 Amp
so I(3 ohm) = I(6 ohm) = 1 Amp
so voltage V(3 ohm) = 1 x 3 = 3 volt
6 ohm is the resultant of parallel combination of 8 ohm and 24 ohm
so voltage across 6 ohm V(6 ohm) = 1 x 6 = 6 volt
voltage is same in parallel combination so
V(8 ohm) = V(24 ohm) = 6 volt
so I(8 ohm) = 6 / 8 = 0.75 Amp
and I(24 ohm) = 6 / 24 = 0.25 Amp
8 ohm is the resultant of series combination of 3 ohm and 5 ohm
current is same in series combination so
I(3 ohm) = I(5 ohm) = 0.75 Amp
so the voltage V(3 ohm) = 0.75 x 3 = 2.25 volt
and V(5 ohm) = 0.75 x 5 = 3.75 volt
3 ohm resistance is the resultant of parallel combination of 4 ohm and 12 ohm
so voltage is same in parallel so
V(4 ohm) = V(12 ohm) = V(3 ohm) = 2.25 volt
so current I(4 ohm) = 2.25 / 4 = 0.5625 Amp
and I(12 ohm) = 2.25 / 12 = 0.1875 Amp
(a) I(3 ohm) = 1 Amp I(24 ohm) = 0.25 Amp I(5 ohm) = 0.75 Amp I(4 ohm) = 0.5625 Amp
I(12 ohm) = 0.1875 Amp
(b) V(3 ohm) = 3 volt V(24 ohm) = 6 volt V(5 ohm) = 3.75 volt V(4 ohm) = 2.25 volt
V(12 ohm) = 2.25 volt