(a) ideal ammeter (Give your answer in mA.) 409.94 ✓ MA (b) ideal voltmeter (Give your answer in volts.) X Look for a path between the two ends of the voltmeter that does not involve any resistance so that currents do not affect the potential differences along that path. V (c) What If? How would the readings in the ammeter (in mA) and voltmeter (in volts) change if the 4.50 V battery was flipped so that its positive terminal was to the right? ideal ammeter X MA ideal voltmeter ✓ V 11.1

C -Q (A) 0.25 uF 10V 2uF 4uF (B) Figure 3: 3. In Figure 3 the capacitors were discharged before being connected to the battery. (a) In Figure 3A the net positive charge delivered by the battery is +Q1 to C1 and a net negative charge -[Q2 + Q3] is delivered to C2 and C3 [-Q2 is delivered to C2 and -Q3 to C3]. From conservation of charge[net charge is zero since both the battery and the capacitors have a zero net excess charge] and path independence of the electric field[sum of voltages must add up to zero for a closed path or conservation of energy]: Q1 + (-Q2) + (-Q3) = 0 conservation of charge +%-왕-용 (True, False) = 0 conservation of energy +V - = 0 conservation of energy (b) If (a) is true then the charge Q1, Q2 and Q3 follow from above three equations as C1(C2 + C3) V Q2 = Ci + C2 + C3 C1 C3 Ci + C2 + C3 Q1 = Q3 = V, (True, False) C1 + C2 + C3 (c) In Figure 3A the net positive charge delivered by the battery is Qtot = Q1 = C1(C2+C3) v which implies an equivalent capacitance of Qtot…

R2 Q1 ... VCC R1 Show your complete solution. Use two decimal places for your final answer. Calculate the value of RC in ohms. at VCC=18 V, IB=26 uA, and VC=7V and B=69

Which of the following statements regarding circuits are true? To find resistance in a parallel circuit, you must sum the individual resitances The direction of conventional current is the direction of positive charge flow. In a circuit, electrons flow from high potential energy (at the negative terminal) to low potential energy (at the positive terminal). When electricity is passed through a lightbulb, the electrons are 'used up' by the bulb. All voltages in a loop sum to 12V The current that flows into a resistor is the same as the current that flows out. To 'short circuit' something means there is an alternate path of less resistance for the current to flow. Switching on a light happens immediately because electrons move from the power source to the bulb incredibly quickly. A 'kilowatt-hour' is a unit of power, not energy.

please show clear step by step working how you got the answer thank you

Zener Diodes: In the following diagram, Iz = 0.5 mA, Vz = 5, rz = 30 ohms. What is Vout? What is the current through D5? Current through D7?

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A student wants to measure the electric potential difderence across the resistor R2 for the circuit below.

What is the total current? + 497 V R4 ko C +0.04 µF L 8 4H 750 Hz O A. 56.6 mA O B. 141 mA O C. 191 mA O D. 244 mA

Please identify the two elements. Choose for the correct answer.