What is a the highest B the lowest total resistance?
Ans: Given resistances R1 =4 Ω, R2=8 Ω, R3=12Ω, and R4=24Ω. ∴ 2 Ω is the lowest total resistance.
How can 3 resistors of resistance?
There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively. This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series. Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω Hence the total resistance of the circuit is 4 Ω.
How will you connect three resistors of 2 ohm?
The 2 ohm and 3 ohm resistors will be connected in series. This will give a resultant resistance of 5 ohm. Then this arrangement will be connected with the 5 ohm resistor in parallel. This will give a effective resistance of 2.5 ohm.
How would you connect 3 resistors to 6 ohms?
Answer. Answer: To obtain a combined resistance of 9 ohms, three resistors each of resistance 6 ohm should be connected in a parallel circuit and one in series. In a parallel circuit, the equivalent resistance of all the resistors connected is equal to the reciprocal value of the sum of the individual resistances.
How would you connect three resistors having values 4 ohm 5 ohm and 6 ohm to get the highest and the lowest equivalent resistance?
51=2ohms and this is added with the resistance of the third resistor in series as 2+4=6 ohms.
How will you join three resistors of resistance 4 ohm 6 ohm and 12 ohm?
1/{1/4+1/6+1/12} = 1/(3/12+2/12+1/12) = 1/(3+2+1)/12 = 1/6/12 = 2 ohm. Two resistors, 20 ohms and 30 ohms, are connected in parallel. This combination is connected in a series to an 8 ohms resistor and a battery of e.m.f. 12 volts.
What would be the highest and the lowest equivalent resistance possible by joining these resistors?
Answer Expert Verified (a) The highest resistance is when the resistances are connected in series. Hence, the highest resistance is 48 Ω. Hence, the lowest resistance is 2 Ω.
How many bulbs of resistance 6 ohm should be joined in parallel to draw a current of 2 ampere from a battery of three Volt?
Answer. Hence 4 bulbs are connected in parallel.
How many bulbs should 8 ohms join?
r = 8 ohm. => n = 4. Therefore 4 resistors (bulb) should be joined in parallel to draw a current of 2A from a battery of 4V.