E = 18 V.
R1 = 61 Ohms
R2 = 34 Ohms
R3 = 90 Ohms
R4 = 11 Ohms
I2 = ?
Req. = R4 + (R1+R2)R3/(R1+R2+R3)
Req. = 11 + (95)90/185 = 11 + 46.22 = 57.22 Ohms.
I4 = E/Req. = 18/57.22 = 0.315A
I2 + I3 = 0.315A
I3 = 95/90 * I2 = 1.056I2
I2 + 1.056I2 = 0.315
2.056I2 = 0.315
I2 = 0.1532A
Four resistors are connected to a battery with a terminal voltage of 18.0 V. The resistors are, R1=61, R2=34, R3=90, and R4=11. R1 & R2 are in series, and combined are in parallel with R3, & the combination of R1, R2, & R3 are in series with R4. Find the current in the 34.0 Ω resistor. Answer in units of A.
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1 answer