# Quick Circuit Practice

1. Jul 30, 2014

### Zondrina

1. The problem statement, all variables and given/known data

What is the emf $(\epsilon)$ of the ideal battery in the following figure?

2. Relevant equations

$i_6 = 1.40 A$
$R_1 = R_2 = R_3 = 2.00 \Omega$
$R_4 = 16.0 \Omega$
$R_5 = 8.00 \Omega$
$R_6 = 4.00 \Omega$

3. 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 $48.3 V$.

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2. Jul 30, 2014

### SammyS

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

Other than that it looks good.