Find the highest and lowest total resistance of a combination of four coils with resistances 4 ohms, 8 ohms, 12 ohms, 24 ohms.

According to the question, there are four coils of resistances = 4 Ω, 8 Ω, 12 Ω and 24 Ω

  1. Given that the coils are connected in series, then equivalent resistance will have to be highest:

⇒ 4 + 8 + 12 + 24 = 48 Ω

  1. Given that the coils are connected in parallel, then equivalent resistance will have to be the lowest:

⇒ 1/R = 1/4 + 1/8 + 1/12 + 1/24

= 1/2

Therefore, R = 2 ohms can be expressed as the least value of resistance which can be obtained from the above 4 resistor combinations.


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  7. X and Y, which are two resistors, with resistances of 2 Ω and 3 Ω respectively are first connected in parallel and then in series. In both cases, the voltage that is supplied is 5 V. (i) Illustrate a circuit diagram to show the combination of resistors. (ii) Calculate the amount of voltage across 3 Ω resistor in the series combination of resistors.
  8. Draw An Electric Circuit With A Cell, Key, Ammeter, A Resistor (Series) Of 2 Ohm With a Combination Of Two Resistors (4 Ohm Each) In Parallel And A Voltmeter Across Parallel Combination.

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