Video: Understanding the Effect of Modifying the Voltage and Resistance in a Circuit

A circuit consists of a battery of voltage ๐‘‰ and a resistor of resistance ๐‘…. The current through the resistor is ๐ผ. The battery is replaced with a battery that has voltage four times as large as the voltage of the original battery and the resistor is replaced with a resistor that has a resistance twice as large as the resistance of the original resistor. What will the current through the new resistor be? [A] ๐ผ [B] 2๐ผ [C] ๐ผ/4 [D] 4๐ผ [E] ๐ผ/2

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Video Transcript

A circuit consists of a battery of voltage ๐‘‰ and a resistor of resistance ๐‘…. The current through the resistor is ๐ผ. The battery is replaced with a battery that has voltage four times as large as the voltage of the original battery. And the resistor is replaced with a resistor that has a resistance twice as large as the resistance of the original resistor. What will the current through the new resistor be? (a) ๐ผ, (b) two ๐ผ, (c) ๐ผ over four, (d) four ๐ผ, (e) ๐ผ over two.

Letโ€™s start off by drawing the original and modified circuits. Here are the two circuits. Both circuits have a battery and a resistor, and the arrow in the middle represents the flow of current. Letโ€™s now label these diagrams with the values we know from the question. In the original circuit, the voltage of the battery is ๐‘‰. So, we put a ๐‘‰ next to the battery to signify that thatโ€™s its voltage. Similarly, the resistance is ๐‘…. So, weโ€™ll label the resistor with the symbol ๐‘…. Finally, the current through the circuit is ๐ผ. So, weโ€™ll label the current arrow that represents the flow about the circuit with the symbol ๐ผ.

In the modified circuit, we know that the voltage of the battery is four times as large as the voltage of the battery in the original circuit. In the original circuit, the battery had voltage ๐‘‰. So, in the modified circuit, the battery has voltage four ๐‘‰. Similarly, the resistor in the modified circuit has resistance twice as large as that of the resistor in the original circuit. In the original circuit, the resistor had resistance ๐‘…. So, in the modified circuit, the resistor has resistance two ๐‘…. What about the current in the modified circuit? Well, the point of the question is to figure out what that current is. So, for now, weโ€™ll just give it the symbol ๐ผ two.

Since the information in the question is specified as voltages, resistances, and currents and what weโ€™re looking for is the current through a resistor, this suggests that we employ Ohmโ€™s law. Ohmโ€™s law provides a relationship between the voltage across a resistor, the current through the resistor, and the value of the resistorโ€™s resistance. Specifically, Ohmโ€™s law says that this relationship takes the form of voltage across the resistor is equal to current through the resistor times resistance of the resistor. Weโ€™ve chosen to write Ohmโ€™s law verbally rather than symbolically just to avoid confusion with the symbols in the question.

And anyway, using Ohmโ€™s law, if weโ€™re looking for one of voltage, current, or resistance, we could solve for it by knowing the other two. Turning to our modified circuit, weโ€™re looking for the value of current, ๐ผ two, through the resistor. So, if we know the resistance and we know the voltage across the resistor, we could use Ohmโ€™s law to find the current. We do know the resistance; itโ€™s two ๐‘…. And since the resistor is the only other component in the circuit besides the battery, the voltage across the resistor is the same as the voltage of the battery, four ๐‘‰. So, using Ohmโ€™s law, we can write the voltage, four ๐‘‰, is equal to the current, ๐ผ two, times the resistance, two ๐‘…. If we divide both sides by two ๐‘…, then on the right-hand side, two ๐‘… divided by two ๐‘… is one. And on the left-hand side, four divided by two is two. And weโ€™re left with ๐ผ two is equal to two times ๐‘‰ over ๐‘….

The trouble is all the answer choices are given in terms of ๐ผ, but we just found the answer in terms of ๐‘‰ over ๐‘…. So, we need some way to relate ๐‘‰ over ๐‘… to ๐ผ. ๐‘‰, ๐‘…, and ๐ผ are the three values from the original circuit. So, letโ€™s use Ohmโ€™s law again to try to find the relationship we need. In this circuit, the resistor has a resistance of ๐‘…. And the current through the resistor is given as ๐ผ. Just like in the modified circuit, the resistor is the only component besides the battery. So, the voltage across the resistor is the same as the voltage of the battery, which is ๐‘‰.

Ohmโ€™s law then gives us ๐‘‰ is equal to ๐ผ times ๐‘…. But remember, the quantity that weโ€™re looking to replace is ๐‘‰ over ๐‘…. So, letโ€™s divide both sides by ๐‘… so that the left-hand side becomes ๐‘‰ over ๐‘…, and we can figure out what this is equal to on the right-hand side. ๐‘… divided by ๐‘… is just one. So, weโ€™re left with ๐ผ. Thus, we find that ๐‘‰ over ๐‘… equals ๐ผ. Using this to replace ๐‘‰ over ๐‘… in our equation for ๐ผ two, we find that ๐ผ two is equal to two ๐ผ. And this is the answer weโ€™re looking for, the current in the modified circuit in terms of the current in the original circuit. So, the correct answer is choice (b), two ๐ผ.

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