2. (a) Two 2. (a) Two resistors in connected in series. The current in one is 0.
ID: 1532055 • Letter: 2
Question
2. (a) Two 2. (a) Two resistors in connected in series. The current in one is 0.350 A. What is -l. resistor the current in the other resistor? Explain your answer. (b) An 9.25 volt battery is connected to two resistons. The potential difference across one resistor is 6.50 volts. What is the potential difference across the other resistor? Are the resistors connected in parallel or in series? Plcase give detailed explanation. (15 pts) 3. Find the equivalent resistance for each of the resistor configurations shown. Please show detailed calculations and two significant figures. (8 pts) 10 oluus 5 ohms 5 ohms 5 olas 10 olams 10 obuus 5 olunas 10 ohms 10 ohms 4. (15 pts A student constructed the circuit shown in the figure. Ru 4.00 2, R 6.00 2 R 3.00 2 and E 36.0 V. (a) Calculate equivalent resistance for this circuit. (b) What are the values for the currents through battery and R3? How much is the potential across R3?(c) What is the potential difference between points a and b? (d) Calculate current through R1? (e) What is the current through resistor Rz? Please first write the values in sambolis orm each part 3. The four identical resistors with resistance 4.50 are connected across a 9.00 v battery as shown in the figure below (a) What is the equivalent resistance for this circuit? (b) If you are to connectammeter right after battery, what value it would read? (c) What is the potential across R1? (d) What is the potential across R3? (e) Calculate the current through each resistor. How the currents are related one to another? Please first write the values in symbolic form for each part. (15 pts)Explanation / Answer
)(a) The current in the other resistor would also be 0.35 A
I = 0.35 A
Because, its a fact that the same current flows though the devices connected in sereis.
(b)V(battery) = 9.25 V
V(other) = 9.25 - 6.5 = 2.75 V
The resistors are connected in series because the potential diffrence is different for two resistors. Its a fact that the resistors connected in parallel have same potential across them.
3)10 and 10 are in parallel and 5 and 5 are in parallel, their combination is in series. So,
Req = 10 x 10 / (10 + 10 ) + 5 x 5 / (5+5) = 5 + 2.5 = 7.5 Ohm
Req = 7.5 Ohm
two 5 are in series, 2 10 are in series, their combination is in parallel and this total combi is in series with 10
R1 = 10*x20/(10+20) = 6.67
Req = 6.67 + 10 = 16.67 Ohm
Req = 16.67 Ohm
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