Question

Consider the following figure. (Assum e R1-29.0 Ω, R2-2 1.0 Ω, and V-16.0 VJ RI 10.0V 5.00 Ω (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 Kirchoff'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 Ω= 5.00 Ω- 29.0 Ω = Need Help? Read it

Answer 1

R_1 = 29.0 Ohm R_2 = 21.0 Ohm V = 16.0 V (b) Let the current in upper loop 1 is I_1 and lower loop is I_2 is the direction as shown in the figure below. Applying KVL in loop (1) and (2) -v + I_1R_1 + 5.0 (I_1 - I_2) + 10 = 0 loop 1 -10 + 5(I_1 - I_1) + R_2 I_2 = 0 loop 2 put the value of V, R_1 and R_2 34 I_1 - 5I_2 = 6 implies I_2 = 34 I_1 - 6/5 -5I_1 + 26I_2 = 10 -5I_1 + 26 (34 I_1 - 6/5) = 10 -25I_1 + 26 times 34 I_1 - 26 times 6 = 50 859 I_1 = 206 implies I_1 = 0.239 A and I_2 = 34 times 0.239 - 6/5 = 0.430 A I_2 = 0.430 A \r\n Hence the cement through R_1 (21 Ohm) = 0.239 A (away from v) 5.00 Ohm = (I_2 - I_1)