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Homework Statement
What is the emf ##(\epsilon)## of the ideal battery in the following figure?
Homework Equations
##i_6 = 1.40 A##
##R_1 = R_2 = R_3 = 2.00 \Omega##
##R_4 = 16.0 \Omega##
##R_5 = 8.00 \Omega##
##R_6 = 4.00 \Omega##
The Attempt at a Solution
We can find the potential right away:
##V_6 = i_6R_6 = (1.40 A)(4.00 \Omega) = 5.6 V##
Note that ##i_5 = i_6## since ##R_5## and ##R_6## are in series, so we can find the other potential:
##V_5 = i_5R_5 = (1.40 A)(8.00 \Omega) = 11.2 V##
Now we note ##V_4 = V_5 + V_6 = 5.6 V + 11.2 V = 16.8 V## since the resistors are in parallel. Now ##i_4## can be found:
##i_4 = \frac{V_4}{R_4} = \frac{16.8 V}{16.0 \Omega} = 1.05 A##
By KCL once again, we have ##i_2 = i_4 + i_5 = 1.05 A + 1.40A = 2.45 A##. So that:
##V_2 = i_2R_2 = (2.45 A)(2.00 \Omega) = 4.90 V##
Now here is where I have a question. Would it be the case that ##R_2## and ##R_4## are in series? Then I would have ##R_3## in parallel and:
##V_3 = V_2 + V_4 = 4.90 V + 16.8 V = 21.7 V##
So that ##i_3 = \frac{V_3}{R_3} = \frac{21.7 V}{2.00 \Omega} = 10.85 A = 10.9 A##.
Appyling KCL again yields ##i_1 = i_2 + i_3 = 2.45 A + 10.85 A = 13.3 A##.
So we get ##V_1 = i_1 R_1 = (13.3 A)(2.00 \Omega) = 26.6 V##.
Appling the loop rule to the leftmost loop, we obtain:
##\epsilon - V_1 - V_3 = 0 \Rightarrow \epsilon = V_1 + V_3 = 26.6 V + 21.7 V = 48.3 V##.
Hence the emf is ##48.3 V##.