Part A = counterclockwise
Part B = I = 2EMF / ((2 * R1) + R2)
Part C = IB = 2*EMF / (R1 + 2*R2)
Part D = 0.064 W
Part E = R1 / R2 = 1
Part F = R2 > R1
In which direction does the current in circuit A flow?
What is the current through the resistor of resistance R2 in circuit A?
Calculate the current IB through the resistor of resistance R2 for circuit B.
What is the power dissipated by the resistor of resistance R2 for circuit A, given that Ε = 10 V, R1 = 300 ohms, and R2 = 5000 ohms?
For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?
Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?
Part A = current
Part B = I2 + I3 – I1
Part C = I3 ⋅ R3 – I2 ⋅ R2
Part D = Vb – I1 ⋅ R1 – I3 ⋅ R3
The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2).
Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.
Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.
Part A = Zero
Part B = Ε
Part C = clockwise
Part D = the top plate
Part E = zero
Part F = q = CΕ
Part G = W = CΕ^2
Part H = q(t) = CΕ * (1 – e^(-t/RC))
Part I = I(t) = (Ε / R) * e^(-t/RC)
Part J = I(t) = (Ε / R) * e^(-t/RC)
Part K = I(t) = -q0 * e^(-t/RC) / (RC)
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Part A = 9.0 v
Part B = 1.6 v
Part C = 0.39 v
Part D = Increase
What does the voltmeter read on a sunny day?
What does the voltmeter read on a cloudy day?
What does the voltmeter read at night?
Does the voltmeter reading increase or decrease as the light intensity increases?
Part A = PR(t) = I(t)2R
Part B = PL(t) = -I(t)2R
Part C = Power comes out of the inductor and is dissipated by the resistor
Consider an R-L circuit with a DC voltage source, as shown in the figure. This circuit has a current Ic when t < 0. At t = 0 the switch is thrown removing the DC voltage source from the circuit. The current decays to I(t) at time t.
What is the power, PR(t), flowing into the resistor, R, at time t?
What is the power flowing into the inductor?
Compare the two equations for power dissipated within the resistor and inductor. Which of the following conclusions about the shift of energy within the circuit can be made?
If the external resistance is then changed to R2 = 4.00 Ω, what is the value of the current I2 in the circuit?Click for More...