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Electrical Circuits

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Submitted By vishal312
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4. Two resistors, R1=8.8 Ω and R2=4.4 Ω, and two uncharged capacitors, C1=0.48 μF and C2 = 0.24 μF, are arranged as shown in Figure 3. A potential difference of 24 V is applied across the combination as shown. a. What is the potential at point a with the switch S open? (Let V = 0 at the negative terminal of the source.) = 8V (voltage divider rule) b. What is the potential at point b with the switch open? = 16V c. What is the charge on C1 with the switch open? = [Q=CV] = 0.48*10-6 *8 = 3.84µC d. What is the potential at a when the switch is closed? =8V e. What is the charge on C1 after the switch is closed? =7.68 µC f. What is the charge on C2 after the switch is closed? =1.82µC

[pic] Figure 3.


A 20.0 m length of wire consists of 10.0 m of copper followed by 10.0 m of aluminum, both of diameter d = 1.6 mm. A voltage difference of 80 mV is placed across the composite wire. Resistivities of Copper and Aluminum are 1.68 x 10-8 Ω.m and 2.82 x 10-8 Ω.m, respectively.

g. What is the total resistance (sum) of the two wires? =0.22378 (using R=ρL/A) h. What is the current through the wire? = 0.357A (using I=V/R)

i. What is the voltage across the copper part? = 29.8mV (voltage divider rule)

j. What is the voltage across the aluminum part? = 50.2mV

k. What is the total electrical power used in the wires? = 0.0285W or 28.5mW (P=I2R)


For the network shown in Figure 4, each of the resistors shown have resistance R = 12.0 Ω. A 12.0-V battery is connected between ends A and B (Assume that end A is connected to the positive terminal of the battery).

l. Find the equivalent resistance between ends A and B.=32.8 ohm

m. The current through the resistor a.=0.366A

n. The current through the resistor c.=0.366A

o. The current through the resistor b. =0.268A

p. The current through the middle (vertical) resistor. =0.073A


Figure 4

7. The circuit in Figure 5 consists of a battery with EMF ε = 3.0 V, a resistor with resistance R = 10 Ω, an ammeter, and a voltmeter. The voltmeter and the ammeter (labeled V and A, respectively) can be considered ideal; that is, their resistances are infinity and zero, respectively. The current in the resistor is I, and the voltage across it is V. The internal resistance of the battery rint = 0.2 Ω. a. What will the ammeter read?= 0.294A b. What is the terminal voltage?= 3V c. What will the voltmeter read? = 2.94V d. What is the power lost in the battery? = 0.0173W e. What is the power PR dissipated in the resistor? = 0.8644W f. What is the total power generated by the battery? = 0.882W [pic] Figure 5

8. For the circuit shown in Figure 6, find q. Equivalent resistance of the circuit.= [(1/820 +1/680)^-1 + 470] = 841.73ohm r. Current through the 470 Ω resistor. = 0.01426 A (using current divider rule) s. Current through the 820 Ω resistor. = 0.006463A t. Current through the 680 Ω resistor. = 0.007795A

[pic] Figure 6

9. For the circuit shown in Figure 7, R1 = R2 = R3 = R4 = 6 Ω, and the battery voltage V = 6 V. u. Find equivalent resistance of the circuit when the switch is open as shown. = 9ohm v. Find current through R1 when the switch is open. = 0.66A w. Find equivalent resistance of the circuit when the switch is closed. = 8ohm x. Find current through R2 when the switch is closed. = 0.25A [pic] Figure 7

10. Refer to the circuit in Figure 8. y. Apply Kirchhoff’s Junction rule at junction a to relate the currents I1, I2, and I3. Use the directions shown in the figure for the currents. = (I3 = I1 + I2) z. Use Kirchhoff’s loop rule to write an equation for the top loop. = (30I1 + 41I3 = 45) aa. Use Kirchhoff’s loop rule to write an equation for the lower loop=(41I3 + 21I2 = 125) ab. Use the three equations in a, b, and c to solve for the magnitudes and directions of the currents I1, I2, and I3.
I1 = -0.858A
I2= 2.583A
I3= 1.725A

[pic] Figure 8

11. A 2.9 kΩ and a 2.2 kΩ resistor are connected in parallel; this combination is connected in series with a 1.3 kΩ resistor. If each resistor is rated at ¼ W (maximum without overheating), ac. What is the maximum current for the circuit? = [I2

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