In figure 27-48, the resistances are R1 = 0.94 and R2 = 2.6 , and the ideal batt

Question

In figure 27-48, the resistances are R1 = 0.94 and R2 = 2.6 , and the ideal batteries have emfs 1 = 2.9 V, and 2 = 3 = 5.4 V. What are the (a) current (in A) in battery 1, (b) the current (in A) in battery 2 and (c) the current (in A) in battery 3? (d) What is the potential difference Va - Vb (in V)? Figure 27-48 (a)NumberUnitsA (b)NumberUnitsA (c)NumberUnitsA (d)NumberUnitsV

Explanation / Answer

Applying Kirchoffs law in Loop 1 2.9 - (IR1) - [(I-I')R2] - 5.4 - IR1 = 0 => 2.9 - 0.94I - 2.6(I-I') - 5.4 - 0.94I = 0 => -2.5 - 6.14I + 2.6I' = 0 => -6.14I + 2.6I' = 2.5          -------eq 1 Applying Kirchoffs law in Loop 2 5.4 + [(I - I')R2] -(I'R1) - 5.4 - (I'R1) = 0 2.6(I - I') -0.94I' - 0.94I' = 0 => 2.6I = 4.48I' => I = 4.48I'/2.6            -------eq2 put this value in eq1 -6.14(4.48I'/2.6 ) + 2.6I' = 2.5 => -10.579I' + 2.6I' = 2.5 => -7.979I' = 2.5 => I' = 0.3133 A => I = 0.5398 A (a) Current in Battery 1 = I = 0.5398 A (b) current in battery 2 = I - I'= 0.2265 A (c)current in battery 3 = I' = 0.3133A (d) Va - [(I-I')R2] - 5.4 = Vb => Va - Vb = 5.4 + .2265*2.6                   = 5.9889 V

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