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The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3$$\Omega$$, 9$$\Omega$$ and 9$$\Omega$$ and a capacitor of 5.0 $$\mu$$F. How much is the current I in the circuit in steady state?
The two batteries are connected in opposition. The net electromotive force ($$\text{EMF}$$) of the loop is $$\epsilon_{\text{net}} = 16.0\text{ V} - 8.0\text{ V} = 8.0\text{ V}$$
$$R_{\text{total}} = 3\ \Omega + 9\ \Omega = 12\ \Omega$$ (in steady state)
$$I = \frac{\epsilon_{\text{net}}}{R_{\text{total}}}$$
$$I = \frac{8.0}{12}$$
$$I = \frac{2}{3}\text{ A} \approx 0.67\text{ A}$$
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