Three resistance coils of 1 Q, 2 Q and 3 Q are connected in series. If the combination is connected to a battery of 9 V, what is the potential drop across the resistance coil of 3 fi ?

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Q: 90 (NDA-II/2011)
Three resistance coils of 1 Q, 2 Q and 3 Q are connected in series. If the combination is connected to a battery of 9 V, what is the potential drop across the resistance coil of 3 fi ?

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,4,10,2,4,4,4

keywords: 

{'resistance coil': [0, 0, 0, 1], 'resistance coils': [0, 0, 0, 1], 'potential drop': [0, 0, 0, 1], 'volt': [0, 0, 0, 1], 'combination': [1, 0, 1, 4], 'battery': [0, 0, 1, 3], 'series': [0, 1, 1, 0]}

In this question, we have three resistance coils connected in series, with resistance values of 1 ohm, 2 ohms, and 3 ohms. When resistance coils are connected in series, the total resistance is equal to the sum of the individual resistances. In this case, the total resistance is 1 + 2 + 3 = 6 ohms.

The potential drop across a resistor in a series circuit is directly proportional to its resistance. This means that the potential drop across the resistor with a resistance of 3 ohms will be greater than the potential drop across the resistors with lower resistances.

To find the potential drop across the resistance coil of 3 ohms, we can use Ohm`s Law. Ohm`s Law states that the potential difference (V) across a resistor is equal to the current (I) through the resistor multiplied by the resistance (R). In this case, the battery has a voltage of 9 volts and the total resistance is 6 ohms, so the current flowing through the circuit is 9V/6Ω = 1.5A.

The potential drop across the resistance coil of 3 ohms can now be calculated as 1.5A * 3Ω =

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