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3 parts
a

503 Ω≈16.67 Ω\frac{50}{3} \, \Omega \approx 16.67 \, \Omega

b

1825 A=0.72 A\frac{18}{25} \, A = 0.72 \, A

c

625 A=0.24 A\frac{6}{25} \, A = 0.24 \, A

Step-by-step solution

5 steps

Here's the solution to your circuit problem:

Part a· 2 steps

  1. Calculate the equivalent resistance of the parallel combination (RpR_p) of R2R_2, R3R_3, and R4R_4.

    1Rp=1R2+1R3+1R4\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} 1Rp=115 Ω+120 Ω+130 Ω\frac{1}{R_p} = \frac{1}{15 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{30 \, \Omega}

    To add these fractions, find a common denominator, which is 60.

    1Rp=460 Ω+360 Ω+260 Ω\frac{1}{R_p} = \frac{4}{60 \, \Omega} + \frac{3}{60 \, \Omega} + \frac{2}{60 \, \Omega} 1Rp=4+3+260 Ω=960 Ω\frac{1}{R_p} = \frac{4+3+2}{60 \, \Omega} = \frac{9}{60 \, \Omega} Rp=609 Ω=203 ΩR_p = \frac{60}{9} \, \Omega = \frac{20}{3} \, \Omega

  2. Calculate the total resistance (RTR_T) of the circuit.

    R1R_1 is in series with RpR_p.

    RT=R1+RpR_T = R_1 + R_p RT=10 Ω+203 ΩR_T = 10 \, \Omega + \frac{20}{3} \, \Omega RT=303 Ω+203 Ω=503 ΩR_T = \frac{30}{3} \, \Omega + \frac{20}{3} \, \Omega = \frac{50}{3} \, \Omega

    The total resistance of the circuit is 503 Ω\frac{50}{3} \, \Omega.

    503 Ω≈16.67 Ω\boxed{\frac{50}{3} \, \Omega \approx 16.67 \, \Omega}

Part b

1825 A=0.72 A\frac{18}{25} \, A = 0.72 \, A

  1. Calculate the total current (ITI_T) flowing from the source using Ohm's Law (I=VRI = \frac{V}{R}).

    IT=VRTI_T = \frac{V}{R_T} IT=12 V503 ΩI_T = \frac{12 \, V}{\frac{50}{3} \, \Omega} IT=12×350 A=3650 A=1825 AI_T = \frac{12 \times 3}{50} \, A = \frac{36}{50} \, A = \frac{18}{25} \, A

    The total current flowing from the source is 1825 A\frac{18}{25} \, A.

    1825 A=0.72 A\boxed{\frac{18}{25} \, A = 0.72 \, A}

Part c

625 A=0.24 A\frac{6}{25} \, A = 0.24 \, A

  1. Calculate the voltage (VpV_p) across the parallel combination (RpR_p).

    This is the voltage across R2R_2, R3R_3, and R4R_4.

    Vp=IT×RpV_p = I_T \times R_p Vp=1825 A×203 ΩV_p = \frac{18}{25} \, A \times \frac{20}{3} \, \Omega Vp=18×2025×3 V=6×2025 V=12025 V=245 VV_p = \frac{18 \times 20}{25 \times 3} \, V = \frac{6 \times 20}{25} \, V = \frac{120}{25} \, V = \frac{24}{5} \, V

  2. Calculate the current (I3I_3) flowing through the 20 Ω20 \, \Omega resistor (R3R_3) using Ohm's Law.

    I3=VpR3I_3 = \frac{V_p}{R_3} I3=245 V20 ΩI_3 = \frac{\frac{24}{5} \, V}{20 \, \Omega} I3=245×20 A=24100 A=625 AI_3 = \frac{24}{5 \times 20} \, A = \frac{24}{100} \, A = \frac{6}{25} \, A

    The current flowing through the 20 Ω20 \, \Omega resistor is 625 A\frac{6}{25} \, A.

    625 A=0.24 A\boxed{\frac{6}{25} \, A = 0.24 \, A}
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