JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
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)
$$I = \frac{\epsilon_{\text{net}}}{R_{\text{total}}}$$
$$I = \frac{8.0}{12}$$
$$I = \frac{2}{3}\text{ A} \approx 0.67\text{ A}$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
