if a 6 volt battery of external resistance 0.5ohms connected to 3 resistors of values 3 ohms,4 ohms,5 ohms.calculate the current in each resistor,the potential difference across the 3ohm resistor and across the parallel resistors.
total resistance R=0.5+3+4×54+5=5.722ΩR=0.5+3+\frac{4\times 5}{4+5}=5.722\OmegaR=0.5+3+4+54×5=5.722Ω
I3Ω=I0.5Ω=6/5.722=1.049A(in series)I_{3\Omega}=I_{0.5\Omega}=6/5.722=1.049A (in \space series)I3Ω=I0.5Ω=6/5.722=1.049A(in series)
V0.5Ω=1.049×0.5=0.5245VV_{0.5 \Omega}=1.049\times 0.5=0.5245VV0.5Ω=1.049×0.5=0.5245V
V3Ω=1.049×3=3.147VV_{3\Omega}=1.049\times 3=3.147VV3Ω=1.049×3=3.147V
V4Ω and 5Ω=IR=1.049×4×54+5=2.33V(in parallel)V_{4 \Omega \space and \space 5 \Omega}=IR=1.049\times \frac{4\times 5}{4+5}=2.33V (in \space parallel)V4Ω and 5Ω=IR=1.049×4+54×5=2.33V(in parallel)
I4Ω=2.33/4=0.5825AI_{4\Omega}=2.33/4=0.5825AI4Ω=2.33/4=0.5825A
I5Ω=2.33/5=0.466AI_{5\Omega}=2.33/5=0.466AI5Ω=2.33/5=0.466A
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