Question

1 Ch 18 Probleris - PHY 1020/109 x C Consider the following Figure 1 x G how lo lake a screenshol - Goog x + - x E → C webassign.net/web/Student/Assignment-Responses/last?dep=22114201#Q4 Use the exact values you enter in previous answer(s) to make later calculation(s). Consider the following figure. (Assume R = 33.00, R2 = 24.00, and V = 24.0 V.) 10.0V 5.002 w R (a) Can the circuit shown above be reduced to a single resistor connected to the batteries? Explain. This answer has not been graded yet. (b) Find the magnitude of the current and its direction in each resistor. 24.0 N = 5.00 1 = 33.0 1 = Need Help? Read It 8:59 PM Type here to search о в е и се 11/20/2019 3

Answer 1

24 =33 A 10.0 V + . 5.00 R2 24 TEIT

a)

We can not reduce the circuit to a single resistor circuit, because resistors are not in parallel or series. so we can not combine them to have a single resistor.

b)

Using KVL in loop ABCDEFA

i₁ R₁ + (i₁ + i₂) R₂ = 24

33 i₁ + 24 (i₁ + i₂) = 24

57 i₁ + 24 i₂ = 24

i₂ = (24 - 57 i₁)/24                                                     Eq-1

Using KVL in loop ABCFA

i₁ R₁ - 5 i₂ + 10 - 24 = 0

33 i₁ - 5 i₂ + 10 - 24 = 0

33 i₁ - 5 i₂ = 14

Using eq-1

33 i₁ - 5 ((24 - 57 i₁)/24) = 14

i₁ = 0.42 A

Using eq-1

i₂ = (24 - 57 i₁)/24

i₂ = (24 - 57 (0.42))/24

i₂ = 0.0025 A

In 24 ohm resistor : i₁ + i₂ = 0.42 + 0.0025 = 0.4225 A

In 5 ohm resistor : i₂ = 0.0025 A

In 33 ohm resistor : i₁ = 0.42 A