Animated Solution for Physics - Current Electricity: The current $I$ drawn from the $5\text{ V}$ source will be

A Circuit Simplification Masterclass

At first glance, the circuit presented in this problem might look like a tangled web of resistors designed to confuse you. With multiple parallel and series-looking connections, it's easy to feel overwhelmed. However, the key to mastering circuit analysis is learning to look past the superficial drawing and identify the underlying electrical topology.

Let's embark on a journey to simplify this circuit and find the total current drawn from the battery.

The most crucial step is to trace the path of the current and identify the main nodes. Let's label the positive terminal of the battery as Node A and the negative terminal as Node B.

When the current leaves Node A, it immediately splits into two paths: one going through a $5\Omega$ resistor and another through a $10\Omega$ resistor. Let's call the ends of these resistors Node C and Node D, respectively.

From Node C, the current can either go through a $10\Omega$ resistor to reach Node B, or it can go through a middle $10\Omega$ resistor to reach Node D. Similarly, from Node D, the current can go through a $20\Omega$ resistor to reach Node B.

If you redraw this structure, placing Nodes A and B on the left and right, and Nodes C and D at the top and bottom, a familiar diamond shape emerges. This is the classic Wheatstone bridge configuration!

Whenever you spot a Wheatstone bridge, your immediate reflex should be to check if it is balanced. A bridge is balanced when the ratio of the resistances in the opposite arms is equal. Let's test our circuit:

For the top arms, the ratio is:

For the bottom arms, the ratio is:

Since both ratios are exactly equal ($\frac{1}{2}$), the bridge is perfectly balanced!

Why is this so magical? In a balanced Wheatstone bridge, the electrical potential at Node C is exactly equal to the potential at Node D ($V_C=V_D$). Because there is no potential difference between these two points, no current will flow through the central $10\Omega$ resistor. It acts as if it's not even there, allowing us to completely remove it from our calculations.

With the central resistor removed, the circuit simplifies dramatically. The current now flows through two distinct, parallel branches:

1. The Top Branch: The $5\Omega$ and $10\Omega$ resistors are now in a simple series connection.

2. The Bottom Branch: The $10\Omega$ and $20\Omega$ resistors are also in series.

Now, we just need to find the equivalent resistance ($R_{eq}$) of these two parallel branches. Using the product-over-sum rule:

The entire complex web of resistors behaves exactly like a single $10\Omega$ resistor!

Finally, we can determine the total current ($I$) drawn from the $5\text{ V}$ source using Ohm's Law ($V=IR$):

By recognizing the hidden symmetry of the Wheatstone bridge, we transformed a potentially tedious Kirchhoff's laws problem into a straightforward mental calculation. Always keep an eye out for these elegant patterns!