What is the emf of the ideal battery in this circuit?

[STEMucator]

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##.

 

[SammyS]

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?...

No. They're not in series, but applying the loop rule will give you ##\ V_3 = V_2 + V_4 \ .##

...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##.

Other than that it looks good.