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- Securecrt & securefx 8 0 2 download free. Let's identify the currents through the resistors by the value of the resistor (I1, I2, I3, I4) and the currents through the batteries by the side of the circuit on which they lay (IL, IR). Start with the 2 Ω resistor. Hyperdock 1 8 0 1. Curio 12 2 – brainstorming and project management app. Apply the loop rule to the circuit on the lower right.
20 V = I2(2 Ω) + (3 A)(4 Ω) I2 = 4 A Proceed to the 3 Ω resistor. Apply the junction rule to the junction in the center of the circuit.I2 = I3 + I4 4 A = I3 + 3 A I3 = 1 A The current through the 1 Ω resistor most certainly runs from right to left. If we apply the loop rule to the top circuit, we'll have to run against that current. This changes what is normally considered a potential drop into a potential increase. (Kind of like skiing up a mountain instead of down.)I1(1 Ω) = (4 A)(2 Ω) + (1 A)(3 Ω) I1 = 11 A - Apply the loop rule to the outer circuit to get the voltage of the battery on the left (continuing with the assumption that the current is running counterclockwise). We find ourselves running through the left battery backwards. This changes what is normally considered a potential increase into a potential decrease. (Kind of like using the ski lift to travel down a mountain instead of up.)
20 V = (11 A)(1 Ω) + VL VL = 9 V Let's verify this result by repeating the procedure for the bottom circuit.20 V = (4 A)(2 Ω) + (1 A)(3 Ω) + VL VL = 9 V Good, we got the same answer using two different approaches. We must be doing the right thing. - The power delivered to the circuit by the battery on the right is the product of its voltage times the current it drives around the circuit. We already have the voltage (it's given in the problem) all that remains is to determine the current. Apply the junction rule to the junction on the left…
IL = I1 + I3 IL = 11 A + 1 A IL = 12 A and again to the junction at the bottom…IR = IL + I4 IR = 12 A + 3 A IR = 15 A to find the power of the battery on the right…P = VI P = (20 V)(15 A) P = 300 W