Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of
$(i)$. $9 \Omega$,
$(ii)$. $4 \Omega$.
Given: Three resistors, each of $6\ Ω$ resistance.
To do: To connect the given three resistors in such a way so that the combination has a resistance of $(i).\ 9\ Ω$, $(ii).\ 4\ Ω$.
Solution: If we connect the three resistors in series, the equivalent resistance will be $=6\ Ω+6\ Ω+6\ Ω=18\ Ω$, that is wrong.
If we connect the three resistors in parallel, the equivalent resistance will be $=\frac{1}{\frac{1}{6\ Ω}+\frac{1}{6\ Ω}+\frac{1}{6\ Ω}}$
$=\frac{1}{\frac{3}{6\ Ω}}$
$=2\ Ω$, that is also not the desired result.
$(i)$. If we connect two resistors in parallel,
Then, equivalent resistance $=\frac{1}{\frac{1}{6\ Ω}+\frac{1}{6\ Ω}}$
$=\frac{1}{\frac{2}{6}}$
$=3\ Ω$
Let's connect the third resistor of $6\ Ω$ in series with the combination of two paralleled resistors$(6\ Ω\ each)$.
Then, $R_{eq}=6\ Ω+3\ Ω=9\ Ω$
$(ii)$. If we connect the two resistors$(6\ Ω\ each)$ in series
Then equivalent resistance $=6\ Ω+6\ Ω=12\ Ω$
Now, let's connect the third resistor$(6\ Ω)$ in parallel with the series combination of two resistors$(6\ Ω\ each)$
Resultant resistance $R_{eq}=\frac{1}{\frac{1}{12\ Ω}+\frac{1}{6\ Ω}}$
$=\frac{1}{\frac{1}{4}}$
$=4\ Ω$
Related Articles
- Compare the power used in the $2 \Omega$ resistor in each of the following circuits:$(i)$ a $6\ V$ battery in series with $1\ \Omega$ and $2\ \Omega$ resistors, and$(ii)$ a $4\ V$ battery in parallel with $12\ \Omega$ and $2\ \Omega$ resistors.
- The resistance of two conductors in parallel is \( 12 \Omega \) and in series is \( 50 \Omega \). Find the resistance of each conductor.
- What is the maximum resistance which can be made using four resistors each of resistance $\frac{1}{2}\Omega $ is -
- A $6\Omega$ resistance wire is double up by folding. Calculate the new resistance of the wire.
- A car headlight bulb working on a $12 \mathrm{~V}$ car battery draws a current of $0.5 \mathrm{~A}$. The resistance of the light bulb is :(a) $0.5 \Omega$(b) $6 \Omega$(c) $12 \Omega$(d) $24 \Omega$
- (a) Three resistors of resistances R1, R2 and R3 are connected (i) in series, and(ii) in parallel. Write expressions for the equivalent resistance of the combination in each case.(b) Two identical resistors of 12 $\Omega$ each are connected to a battery of 3 V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.
- If a person has five resistors each of value $\frac {1}{5}\Omega$, then the maximum resistance he can obtain by connecting them is(A) 1W (B) 5W (C) 10W (D) 25W
- A heater of resistance $50\ \Omega$ is connected to the $220\ V$ line. How much current will the heater draw?
- Explain with the help of a labeled circuit diagram how you will find the resistance of a combination of three resistors, of resistance R1, R2 and R3 joined in parallel. Also mention how you will connect the ammeter and the voltmeter in the circuit when measuring the current in the circuit and the potential difference across one of the three resistors of the combination.
- Two identical resistors, each of resistance 15 W, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case.
- Name the electrical property of a material whose symbol is omega"."
- How many $176\ \Omega$ resistors $(in\ parallel)$ are required to carry $5\ A$ on a $220\ V$ line?
- How to find the resistance of 'n' number of resistors connected in parallel combination?
Kickstart Your Career
Get certified by completing the course
Get Started