Question

SOLVE THIS PLOX!!
IT'S URGENT, DON'T SPAM
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Answer 1

$\underline{ \sf {\maltese \; \; \; Question : \; \; \; }}$

Show that how you would connect three resistors, each of resistance 6Ω so that combination has a reactance of,

• (a) 9Ω
• (b) 4Ω

$\underline{ \sf {\maltese \; \; \; Solution : \; \; \; }}$

Note :

Whenever the resistors are connected in series combination, then equivalent resistance is given by,

$\boxed{\sf{ R_S = R_1 + R_2 + \dots R_n}}$

Whenever the resistors are connected in parallel combination, then equivalent resistance is given by,

$\boxed{\sf{ \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \dots + \dfrac{1}{R_n} }}$

Explication of steps :

To get the equivalent resistance of 9Ω :

Let,

• $\sf R_1$ = 6Ω
• $\sf R_2$ = 6Ω
• $\sf R_3$ = 6Ω

Now, if we connect $\sf R_1$ and $\sf R_2$ in a parallel combination and the combined resistance of $\sf R_1$ and $\sf R_2$ be connected in series combination with $\sf R_3$, then the equivalent resistance will become 9Ω.

Verification :

As, t $\sf R_1$ and $\sf R_2$ are connected in parallel combined, so their combined resistance will be given by,

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2)}} = \dfrac{1}{6} + \dfrac{1}{6} \; \Omega}$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2)}} = \dfrac{1 + 1}{6} \; \Omega}$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2)}} = \dfrac{2}{6} \; \Omega}$

On reciprocating both sides,

$\quad \twoheadrightarrow\sf { R_{(1,2)}= \dfrac{6}{2} \; \Omega}$

$\quad \twoheadrightarrow\underline{\bf { R_{(1,2)}= 3\; \Omega}}$

Now, the combined resistance of $\sf R_1$ and $\sf R_2$ is in series combination with $\sf R_3$ . So, equivalent resistance will be given by,

$\quad \twoheadrightarrow\sf { R_{(1,2,3)} = R_{(1,2)} + R_3}$

$\quad \twoheadrightarrow\sf { R_{(1,2,3)} = (3 + 6) \; \Omega}$

$\quad \twoheadrightarrow\underline{\boxed{\bf { R_{(1,2,3)} = 9 \; \Omega}}}$

To get the equivalent resistance of 4Ω :

Let,

• $\sf R_1$ = 6Ω
• $\sf R_2$ = 6Ω
• $\sf R_3$ = 6Ω

Now, if we connect $\sf R_1$ and $\sf R_2$ in a series combination and the combined resistance of $\sf R_1$ and $\sf R_2$ be connected in parallel combination with $\sf R_3$, then the equivalent resistance will become 4Ω.

Verification :

As $\sf R_1$ and $\sf R_2$ are connected in series combination, so combined resistance of $\sf R_1$ and $\sf R_2$ will be given by,

$\quad \twoheadrightarrow\sf { R_{(1,2)} = R_1 + R_2}$

$\quad \twoheadrightarrow\sf { R_{(1,2)} = (6 + 6) \; \Omega}$

$\quad \twoheadrightarrow\underline{\bf { R_{(1,2)}= 12\; \Omega}}$

Now, as the combined resistance of $\sf R_1$ and $\sf R_2$ is in parallel combination with $\sf R_3$ . So, equivalent resistance will be given by,

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} }$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2,3)}} = \dfrac{1}{12} + \dfrac{1}{6} \; \Omega}$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2,3)}} = \dfrac{1 + 2}{12} \; \Omega}$

$\quad \twoheadrightarrow\sf { \dfrac{1}{R_{(1,2,3)}} = \dfrac{3}{12} \; \Omega}$

On reciprocating both sides,

$\quad \twoheadrightarrow\sf { R_{(1,2,3)}= \dfrac{12}{3} \; \Omega}$

$\quad \twoheadrightarrow\underline{\boxed{\bf { R_{(1,2,3)} = 4 \; \Omega}}}$

Hence, verified!

Answer 2

The only weapon you need to kill this question is the idea about resistors in parallel and in series. All you need to keep an eye on is the resistance they are asking you to obtain (in this question 9 ohm and 4 ohm) and keep trying various ways with the resistor specified in the question (in this case 6 ohm).

If the resistors are put in parallel the equivalent resistance decreases whereas in series the equivalent resistance increases.

If resistors are in parallel,

• Req = 1/R1 + 1/R2 + 1/R3...+ 1/Rn

If resistors are in series,

• Req = R1 + R2 + R3...+Rn

So, with this let's try solving the question.

(a) : 9 Ω

Now if you feel like connecting the 6 Ω in series, you should put your pen down and think again (if you have a brain,ik you don't ._. xD xD)

Because if we decide to put the 6 Ω in series the equivalent resistance we will get is 18 Ω which is of course exceeding the required resistance i.e 9 Ω. So, since we want to obtain a lesser value, the thing you should try to do is put them in parallel. Now it's actually always better if you have less number of resistors in question (in this question we have three) than just try first taking two resistor leaving the third one. So,

1/Req = 1/6 + 1/6

1/Req = 6 + 6 / 36

1/Req = 12/36

1/Req = 1/3

Req = 3 Ω

Now, you should a smile on your face because all you have to do is now put a 6 Ω resistor in series with the 3 Ω resistor. So,

Req = 3 + 6

Req = 9 Ω

So,we are done with the first part. Put two 6Ω resistor in parallel and the remaining one in series with them.

(b) : 4 Ω

So, this question has a value which is already lesser than the value we have in question (6).

So,to decrease the resistance we need to make use of the parallel combination but since we already calculated above the value of resistance in parallel combination of two resistors which was 3 Ω,so if we make use of the same logic,we will need just a 1 Ω resistor which isn't possible to obtain using any combination,so we need to try the series combination because even if the resistance increase overall..we can put it into parallel with other resistors and reduce it but if once the overall decreases...we can't do much with it because it tells us that there is no any such value available so far to make the possible combination exist.

Req = 6 + 6

Req = 12 Ω

Now, since this is much higher value as per our need,we will put this in parallel with another 6Ω resistor to decrease the value.

1/Req = 1/6 + 1/12

1/Req = 6 + 12/ 72

1/Req = 18/72

1/Req = 1/4

$\sf{Req = 4}$

So,here we directly have the required value of resistance. So,put two 6 Ω resistors in series and one 6 ohm in parallel with them.