Animated Solution for Physics - Current Electricity: The current $i$ in the circuit (see figure) is

Welcome to this classic circuit problem! At first glance, the circuit might look a bit tricky with that triangular shape on the right.

However, our goal is simple: we need to find the total current $i$ flowing out of the $2\text{ V}$ battery.

To do this, we must find the equivalent resistance of the entire network. The golden rule for simplifying circuits is to start from the end furthest away from the voltage source.

Let's focus our attention on the rightmost part of the circuit. Notice the diagonal $30\Omega$ resistor and the horizontal $30\Omega$ resistor at the bottom.

If you trace the path, you'll see they are connected end-to-end. There are no other wires branching off from their junction. This means they are strictly in series.

Since they are in series, finding their combined resistance is straightforward. We simply add their values together:

Now, let's look at the vertical $30\Omega$ resistor. This vertical resistor and our newly found $60\Omega$ branch are connected across the exact same two nodes—one at the top and one at the bottom.

To find the total equivalent resistance of these parallel branches, we can use the handy product-over-sum shortcut:

Substituting our values into the formula, we get:

This entire complex resistor network simplifies beautifully to just a single $20\Omega$ resistor!

Ohm's Law

Now we are in the final stretch. We have a simple circuit with a $2\text{ V}$ battery and a total equivalent resistance of $20\Omega$.

We apply Ohm's law, which states that the total current is equal to the total voltage divided by the equivalent resistance:

Substituting our known values into Ohm's law, we have:

This simplifies perfectly to our final answer:

So, the current flowing in the main circuit is $\frac{1}{10}\text{ A}$, which perfectly matches option (c).