Animated Solution for Physics - Current Electricity: The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15\Omega$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Welcome to one of the most classic challenges in circuit analysis: the unbalanced Wheatstone bridge. At first glance, this circuit looks like a nightmare. We have a $10\text{ V}$ battery driving current through a network of resistors, and right in the middle, bridging the gap between nodes B and D, sits a galvanometer with a resistance of $15\Omega$.

If the bridge were balanced—meaning the ratio of the top resistors equaled the ratio of the bottom resistors—we could simply declare the galvanometer current to be zero and move on. However, a quick check reveals that $100/10 eq 60/5$. The bridge is unbalanced.

While we could use Kirchhoff's Voltage and Current Laws to set up a system of three simultaneous equations, that approach is mathematically tedious and highly prone to algebraic errors. Instead, we will use a much more elegant tool: Thevenin's Theorem.

Thevenin's Theorem

Thevenin's Theorem allows us to take any complex linear circuit and simplify it down to a single voltage source ($V_{th}$) and a single series resistor ($R_{th}$) connected to our load. In this case, our "load" is the galvanometer.

To begin, we must temporarily remove the galvanometer from the circuit. This creates an open circuit between terminals B and D. Our goal now is to find the potential difference between these two nodes, which will be our Thevenin voltage, $V_{th}$.

With the galvanometer removed, the circuit beautifully splits into two independent voltage dividers connected in parallel across the $10\text{ V}$ source.

Let's find the potential at node B first. The top branch consists of a $100\Omega$ resistor and a $10\Omega$ resistor in series. The voltage at node B is simply the voltage drop across the $10\Omega$ resistor. Using the voltage divider rule:

Similarly, the bottom branch consists of a $60\Omega$ resistor and a $5\Omega$ resistor. The voltage at node D is the voltage drop across the $5\Omega$ resistor:

Now, the Thevenin voltage is the difference between these two potentials:

Taking the common denominator of $143$, we get:

Next, we need to find the Thevenin resistance, $R_{th}$. The rule here is to turn off all independent sources. For a voltage source, "turning it off" means replacing it with a short circuit.

Imagine a wire connecting node A directly to node C. Now, look into the circuit from terminals B and D. The $100\Omega$ and $10\Omega$ resistors are now connected in parallel between node B and the common A/C node. Likewise, the $60\Omega$ and $5\Omega$ resistors are in parallel between node D and the common A/C node. These two parallel pairs are in series with each other.

Taking the common denominator of $143$ again:

We have successfully reduced the entire bridge to a simple Thevenin equivalent circuit: a voltage source of $\frac{20}{143}\text{ V}$ in series with a resistance of $\frac{1960}{143}\Omega$.

Now, we reconnect our $15\Omega$ galvanometer to this simplified circuit. The current flowing through it is given by Ohm's Law:

To simplify, multiply the numerator and the denominator by $143$:

Dividing by $5$, we get:

Converting this to milliamperes, we arrive at our final answer:

By leveraging Thevenin's Theorem, we bypassed the need for complex simultaneous equations and arrived at the precise answer through a series of logical, manageable steps. Always look for these elegant shortcuts in circuit analysis!