4. 2.00 ΩΣ 2.00 23 a 1.00 2 12.0 V 1.00 Ω 10.0 V ww+ 1.00 Ω 3.00 1.00 2 8.0 V • 2.00 Ω Ω -ww (a) Find the potential Vab of point a with respect to point b. (b) Points a and b are now connected by a resistanceless wire. What is the current in the 12.0 V battery after the wire in connected?
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- Find the equivalent resistance between points A and B in the drawing. Assume R1 =2.70 Q, R2= 6.50 Q, R3 = 1.40 Q, R4 = 3.80 O, R5 = 4.70 Ω, R-2.90 Ω, and R,- 5.00 Ω. R1 R2 R3 A ww Rs R- B ww- R4· (a) In the figure, determine the potential difference Vad. (b) If a battery with emf 9.30 V and internal resistance of 1.00 ohm is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00- V battery. What is now the potential difference Vbc? 0.50 2 4.00 V 9.00 N 6.00 Ω 0.50 N 8.00 V ww 8.00 ΩPLEASE ANSWER ALL PARTS!!!!! VERY URGENT!!!!!!!!!!!!
- In the figure the ideal battery has emf = 31.7 V, and the resistances are R1 = R2 = 41 Ω, R3 = R4 = R5 = 8.2 Ω, R6 = 5.4 Ω, and R7 = 4.8 Ω. What are currents (a)i2, (b)i4, (c)i1, (d)i3, and (e)i5?If a voltmeter that is ideal is connected to measure the voltage across the 83.0-kΩ resistor, what is its reading? Enter the absolute value of the reading. (in image, x = 8.20 V)A 36 V battery is connected to a network of resistors with R1 = 45 Ω, R2 = 36 Ω, and R3 = 108 Ω. What is the potential difference across R3? (in V) (a) 54.0 V (b) 13.5 V(c) 27.0 V(d) 36.0 V
- (a) Given a 52.0 V battery and 20.0 Ω and 84.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I20.0 Ω= A P20.0 Ω= W I84.0 Ω= A P84.0 Ω= W (b)Repeat when the resistances are in parallel. I20.0 Ω= A P20.0 Ω= W I84.0 Ω= A P84.0 Ω= WA battery has ɛ= 30 V, and an internal resistance of 2.2 Q. What is the terminal voltage when it is connected to R= 10 0? а. 8.20 V b. 16.39 V с. 30.00 V d. 12.30 V е. 24.59 V