Question

Use the exact values you enter in previous answer(s) to make later calculation(s). Consider the following figure. (Assume R, = 28.02, R2 = 21.0.2, and V = 22.0 V.) awn 1 + R 10.0 V 5 .00 22 R2 w (a) Can the circuit shown above be reduced to a single resistor connected to the batteries? Explain. No. This multi-loop circuit does not contain any resistors in series (i.e., connected so all the current in one must pass through the other) nor in parallel (connected so the voltage drop across one is always the same as that across the other). Thus, this circuit cannot be simplified any further, and Kirchhoff's rules must be used to analyze it. Score: 1 out of 1 Comment: (b) Find the magnitude of the current and its direction in each resistor. 21.0.2 = 5.00 2 = 28.0 2 = Need Help? Read It

Answer 1

22 I952 ing kirchhoffis gule (Junction Ruled Toy SR I - 1 + 1 - 0 "212 it, +1) = I Applying voltage law on first Loop 22-28 I, + 5 I, 40=0 SI ₂ - 28 I,= -12 281-512 = 12 - on 2nd loop 10-5 IL- 21 ( I + I₂ ) = 0 10-5 I2- 211, -21I₂ = 0 -21 I, - 26 I 2 =-10 211, +2612 = 10 - multiply eg ② by 26 and eq. ( by s e 7281,-130 I 2 = 312 1051, +130 I2 = 50 @ Adding eq@ to =) = 0.4345738 833 I, to=362 I= 362 20.435A 833 2800.43S-SI2 = 12 - put I, in eg. 12.18-5 1 2 =12 I2 = 0.18 = 0.035 A from ea. O I - Ist I2 = 6.435 +0.036 I= 0.471A So current in 21.00 current in Soor Current in 28.2 I = 0.471 A/ I2 = 0.036A) The direction of current are some that have assumed. I. 5.435 AL