Two resistances of 6 Ω and 12 Ω are connected in parallel. Their net resistance is ______?
Answer: 4 Ω
Explanation
When two resistances are connected in parallel, the net resistance (Rnet) is calculated using the formula:
1/Rnet = 1/R1 + 1/R2
where R1 and R2 are the individual resistances.
In this case, R1 = 6 Ω and R2 = 12 Ω, so:
1/Rnet = 1/6 + 1/12 = 1/4
Rnet = 4 Ω
Therefore, the net resistance of the two resistances connected in parallel is 4 Ω.
This question appeared in
Subjects (4 times)
EVERYDAY SCIENCE (2 times)
GENERAL SCIENCE (2 times)
Related MCQs
- A wire has a resistance of 6 Ω. It is cut into two parts and both the halves are connected in parallel. The new resistance is:
- Two resistance of 2Ω each are connected in parallel to each other resultant resistance will be _____?
- When resistances are connected in series, the current passing through them is _____?
- A resistance R is placed in parallel with another resistance 40Ω their equivalent resistance is 24Ω the value of R is?
- In an induction motor connected in parallel, load can be shifted through ______?
- A resistor of 4kΩ with tolerance 10% is connected in parallel with a resistor of 6 kΩ with tolerance 10%. The tolerance of the parallel combination is nearly ______?
- Two identical capacitors, each with capacitance C, are connected in parallel and the combination is connected in series to a third identical capacitor. The equivalent capacitance of this arrangement is _____?
- Which of the following cannot be connected in parallel because KVL (Kirchhoff's Voltage Law) is not valid?
- Four resistors, each with a value of 68 ohms, are connected in parallel across a 220 V source. What is the total current (I)?
- Two bulbs are connected in a parallel circuit. One bulb blows (blast), what will happen to the other bulb?
