In the circuit shown in the figure, the voltmeter reading is four volts. What is the current through the six 𝛺 resistor?
Looking at the figure, we see that this circuit consists of a power supply, a two 𝛺 resistor, and after that a six and a four 𝛺 resistor arranged in parallel. We want to know how much current moves through the six 𝛺 resistor. We’ll call that 𝐼 sub six 𝛺.
To start doing that, we’ll solve for the total current that runs through the circuit. Ohm’s law will help us solve for this current. This law tells us that the potential difference across an entire circuit or a part of the circuit is equal to the current that runs through that piece multiplied by its resistance.
In our case, we can say that the voltmeter reading — the potential difference across the two 𝛺 resistor — is equal to the total current in the circuit and 𝐼 sub 𝑇 multiplied by two 𝛺. This implies that 𝐼 sub 𝑇 is equal to four volts divided by two 𝛺 or two amps. It’s this two amps of current which moves past our two 𝛺 resistor and then divides up over the two branches of our parallel circuit section.
To find out how much of the current goes to the four 𝛺 resistor branch versus the six 𝛺 resistor branch, we can solve for the total potential difference across the entire circuit. And to do that, we want to know the total equivalent resistance of this circuit. We can write that 𝑅 sub 𝑇 the total circuit resistance is equal to two 𝛺 plus the equivalent resistance of the two resistors arranged in parallel. We’ve called that 𝑅 sub 𝑃.
When two resistors are arranged this way in parallel, their equivalent resistance together is equal to their product divided by their sum. This means that 𝑅 sub 𝑃 in our case is six 𝛺 times four 𝛺 divided by six 𝛺 plus four 𝛺. This is 24 𝛺 over 10 or 2.4 𝛺. And all this implies that our total resistance 𝑅 sub 𝑇 is, therefore, equal to 4.4 𝛺.
Knowing the total current and resistance of this circuit, we can now reapply Ohm’s law to solve for the total potential difference across it. Two amps times 4.4 𝛺 is equal to 8.8 volts, the total potential difference across the entire circuit. We’re told that the voltmeter measuring the potential drop across the two 𝛺 resistor records a value of four volts. That means there are 4.8 volts of potential difference remaining to drop over the parallel branch.
And since current can travel through either one of these two branches, that means that the potential drop across both branches of the parallel circuit is 4.8 volts. Applying Ohm’s law one last time then, we can say that 4.8 volts the potential drop across the six 𝛺 resistor is equal to the current through that resistor times its resistance.
Rearranging, we see 𝐼 sub six 𝛺 is equal to 4.8 volts divided by six 𝛺, which equals 0.8 amps. That’s the current running through the six 𝛺 resistor.