11.
Resistances in series : The equivalent resistance of the circuit is found by simply adding up the resistance values of the individual resistors. Current flowing through each resistance is same.
Resistances in parallel : The equivalent resistance is given by
. Voltage across each resistor is same.
We will also be using ohm's Law :
R2 and R3 are in parallel :
Also, R4, R5 and R6 are in parallel
Now, R1 Req1, Req2, R7 and R8 are in series.
Let Power dissipated, Current and Voltage across each resistor be denoted by the same subscript to avoid confusion eg. current through Req1 is Ieq1 and voltage across it is Veq1.
Now, we can imagine as our circuit consists of one resistor Req and a voltage source of 110V. Therefore,
Since, current flowing through resistances in series is the same.
Hence
Voltage across each resistor :
Since, voltage across resistances in parallel is same.
Hence,
Current flowing through each resistor is given by,
Since, resistances R4, R5 and R6 are equal and voltage across them is also equal.
Hence, Current flowing through them will also be equal i.e. I4 = I5 = I6 = 0.11A
Power dissipated by any resistor can be found using formula :
Power dissipated by resistor = current flowing through it Voltage across it
Since, current and voltage across R4, R5 and R6 are equal hence, power dissipated is also equal i.e. 0.73 W.
Total power dissipated by the circuit is the sum of all the dissipated power
12.
We will be using,
Kirchhoff's Current Law(KCL): Algebraic sum of all the currents at a junction is zero.
Sign Convention : Take currents coming into the junction as positive and currents leaving the junction as negative.
Kirchhoff's Voltage Law (KVL): The algebraic sum of all voltages within the loop must be equal to zero.
Sign Convention : If potential increases during traversing any element, put a positive sign in the voltage drop. If potential decreases during traversing any element, put a negative sign in the voltage drop.
Always remember that potential decreases in the direction of flow of current.
Applying KVL in loop abcdea, (in clockwise direction)
Applying KVL in loop aefgha, (in clockwise direction)
Also, applying KCL at junction a,
I 1 = I 2 + I 3 .................(3)
We have 3 equations and 3 variables, we can easily solve them using elimination method.
I2 flows from b to a not a to b.
Putting I 2 and I 3 in equation (3)
I 1 = -3.5A + 8.25A = 4.75A
Voltage across each resistor can be found using ohm's law:
1000 11. Determine the equivalent resistance of the circuit. Then determine the current in each path,...
3. Determine the equivalent resistance of the circuit. Then determine the current in each path, and the voltage across each resistor. Then determine the power dissipated by each resistor and the total power dissipated by the circuit 10.02 W 4.000 w 5.000 w 2.000 3.000 w 8,00 V 4. Determine the each of the currents in each path, and the voltages across each resistor. E = 24.0 V 0.1002 R 5.0 22 RS 2022 E2 = 48.0 V RE 17,...
3. Determine the equivalent resistance of the circuit. Then determine the current in each path, and the voltage across each resistor. Then determine the power dissipated by each resistor and the total power dissipated by the circuit. 10.0 12 W 4.000 5.00 2.00 3.002 8.00 V HE
20. (a) Determine the equivalent resistance of the circuit in ▼Fig. 18.34 Find (b) the current in each resistor, (c) the voltage across each resistor, and (d) the total power delivered to the circuit. FIGURE 18.34 Power dissipation See Exercise 20. 10Ω 2.0 Ω 6.0 92 V 24 V 4,0Ω 12Ω 10Ω 5.0 Ω 시 r 20. (a) Determine the equivalent resistance of the circuit in ▼Fig. 18.34 Find (b) the current in each resistor, (c) the voltage across each...
a) Calculate the equivalent resistance of this circuit. (1) b) Calculate the current flowing through each resistor. (3) c) Calculate the power dissipated by resistor R3. (1) R2 R3 4012 2012 Ri 302 10 V
8. Suppose the voltage output of a battery is 10.0V in a series circuit, and the resistances are Ri=4.0012, R2=5.0092, and R3=6.002. a. Draw a diagram of the circuit, and determine the total resistance, the current through each resistor, the voltage drop across each resistor, and the power dissipated by each resistor and the source. 9. Suppose the voltage output of a battery is 10.0V in a parallel circuit, and the resistances are Ri=4.0012, R=5.0002, and R3=6.092. b. Draw a...
12. For the following circuit: a) Find the total resistance? b) For each resistor select an appropriate current direction and find the current? c) Find the total current In? d) Show the voltage polarities across each resistor? e) Find the voltage drop across each resistor? f) How much is Vab? g) Calculate the total dissipated power by the circuit? R1 w 200 R3 w 1250 SR4 2000 R5 500 -120 V SR2 1500 a R6 1500 SR7 3000
For the given circuit, find the following: The total load resistance of the circuit between circuit points a and b. The current through each resistor. The voltage across each resistor. The power dissipated by each resistor. The power supplied by the battery. The terminal voltage, Vab. Please be as detailed as possible and explain each step. 10.0 Ω - 8.0 Ω 6.0 Ω 4.0 Ω LW 5.0 Ω a r=0.50 Ω - L (a) & = 9.0V
1801 w 390 M 4. For the following circuit: a) Find the total resistance? b) Find the currently to l? c) Find the current ? d) Find the voltage drop across cach resistor? e) How much is Vab and V? ) Calculate the total dissipated power by the circuit? 1002 560 220 20 V 150 910 w b.
1) what is the equivalent resistance of this circuit? 2) what is the current through each resistor? 3) what is the voltage across each resistor? 1. What is the equivalent resistance of this circuit? 2. What is the current through each resistor? 3. What is the voltage across each resistor? R1 -WE Vbat = 24V + R2 3 twt twt 3 R3 -W- R4 (A) V) TR() RI 2 R2 6 R3 1 R44 V- IR A V... = 0
- + A Read aloud 4. Determine the each of the currents in each path, and the voltages across each resistor. 'E, = 24.0 V wo 0.10.2 R 5.0.2 R. 2022 TA E2 = 48.0 V k 0.500 R w e 40 12 20.20 -0.2012 78 E, 6.0V 36.0V i 0.05 h CourseBanner ng