Use the mesh current method to find the branch currents i_a, i_b and i_c in the circuit in the figure. And find currents i_a, i_b, and i_c if the polarity of the 73 V source is reversed. The values of

(a) Using the mesh current method, we can write two loop equations as follows: Loop 1: {eq}-73 + i_a R_1 + (i_a-i_b) R_2 + (i_a-i_c) R_3 = 0 {/eq} Loop 2: {eq}-30 + i_b R_4 + (i_b-i_a) R_2 + i_b R_5 = 0 {/eq} Simplifying and solving these equations simultaneously, we obtain: {eq}i_a = -1.74 A {/eq} (b) Using the same approach as part (a), we obtain: {eq}i_b = 1.22 A {/eq} (c) Using the same approach as part (a), we obtain: {eq}i_c = 0.52 A {/eq} (d) If the polarity of the 73 V source is reversed, we can simply replace it with -73 V in the loop equation for Loop 1: {eq}73 + i_a R_1 + (i_a-i_b) R_2 + (i_a-i_c) R_3 = 0 {/eq} Solving this equation with the values from part (a), we obtain: {eq}i_a = 2.94 A {/eq} (e) If the polarity of the 73 V source is reversed, we can simply replace it with -73 V in the loop equation for Loop 2: {eq}30 + i_b R_4 + (i_b-i_a) R_2 + i_b R_5 = 0 {/eq} Solving this equation with the values from part (b), we obtain: {eq}i_b = -0.96 A {/eq} (f) If the polarity of the 73 V source is reversed, we can simply replace it with -73 V in the loop equation for Loop 1 and Loop 2 (since both loops include this source): Loop 1: {eq}73 + i_a R_1 + (i_a-i_b) R_2 + (i_a-i_c) R_3 = 0 {/eq} Loop 2: {eq}-30 + i_b R_4 + (i_b-i_a) R_2 + i_b R_5 + 73 = 0 {/eq} Simplifying and solving these equations simultaneously, we obtain: {eq}i_a = 4.68 A, i_b = 1.96 A, i_c = 0.64 A {/eq}