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We know that when a resistor is connected in series , the current flowing through that resistor will be constant.But how do we use a resistor to limit the current flowing through a circuit ,when resistor is connected in series with the circuit ?

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Resistance can be interpreted in various ways depending on the circuit. It can be used to cause a potential drop, or it can be used as a heating device, etc.

You are asking how resistance can change the current flowing through the circuit when connected in series. In that context, the resistance can be used to alter the total resistance of the circuit which affects the current flowing through the circuit.

Ohm's law states that for small voltages (up to 1000V - depends on the device actually), ohmic devices obey the following relation $$I = \frac{V}{R}$$

For a simple circuit such as the one given below, the total resistance decides the amount of current flowing through the circuit.

enter image description here

Here the E.M.F (voltage of the source in an ideal battery) is fixed and what the turns up in the denominator decides how much current will flow through the circuit.

Real-life Application (Plugging an LED to your 12V battery)

Let's talk about a more practical application, say you have an LED bulb of which has a resistance of $2 \Omega$. If you plug it to your 12V battery, the LED is going to blow up because it would draw nearly 6A which is WAY too much.

So does this mean you can't use LED with your battery? No. You can use by connecting a resistor in series with the LED.

The LED needs just a little bit of current, let's say it is rated to operate at 1A of current.

enter image description here

You want the total current flowing through your circuit to be 1A. The total resistance you'll need in the circuit will be given by,

$$R = \frac{V}{I} = \frac{12V}{1A} = 12\Omega$$

$$R = 2\Omega(LED) + R_{resistor}$$ $$R_{resistor} = 10\Omega$$

So you will have to connect a resistor with the resistance of $10\Omega$ in series with the LED to ensure that nearly $1A$ of current passes through the LED.

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    $\begingroup$ Power dissipation on resistor will be 108 W vs LED power 2 W. Resistor will be hot - be careful if you are about to try that at home. $\endgroup$ – lowtech Jun 22 '16 at 21:34
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    $\begingroup$ Is it typo: "to ensure that nearly 2A of current passes through the LED"? It should be exactly 1 A? $\endgroup$ – lowtech Jun 22 '16 at 21:36
  • $\begingroup$ Thanks! You can also use multiple resistors in series so that they don't get burnt due to over-heating. $\endgroup$ – Yashas Jun 23 '16 at 3:38
  • $\begingroup$ Moreover, the electricity supplied through sockets are alternating current. The LED will blow up in this case too. I'd need to add a Zener circuit/diode to reduce the variations. Anyway, I changed the application. $\endgroup$ – Yashas Jun 23 '16 at 3:50

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