Question

Two identical resistors, each of resistance 12Ω, are connected in (i) series (ii) parallel, in turn to a battery of 6V. calculate the ratio of the power consumed in the combination of resistors in each case.

Answer 1

Given :

▪ Two identical resistors each of resistance 12Ω are connected in

• series connection
• parallel connection

▪ Voltage of battery = 6V

To Find :

▪ Ratio of the power consumed in the combination of resistors in each case.

Solution :

→ First we have to find out equivalent resistance for each case.

→ After that we can easily calculate power consumed in the combination of resistors.

✴ Eq. resistance (series) :

☞ Rs = R1 + R2 + ... + Rn

✴ Eq. resistance (parallel) :

☞ 1/Rp = 1/R1 + 1/R2 + ... + 1/Rn

____________________________

▶ First case :

▪ Equivalent resistance :

✒ Rs = R1 + R2

✒ Rs = 12 + 12

✒ Rs = 24Ω

▪ Power consumed :

✏ P = V^2/Rs

✏ P = (6×6)/24

✏ P = 6/4

✏ P = 1.5watt

▶ Second case :

▪ Equivalent resistance :

✒ 1/Rp = 1/R1 + 1/R2

✒ 1/Rp = 1/12 + 1/12

✒ Rp = 12/2

✒ Rp = 6Ω

▪ Power consumed :

✏ P' = V^2/Rp

✏ P' = (6×6)/6

✏ P' = 6watt

____________________________

✴ Taking ratio of P and P', we get

⏭ P/P' = 1.5/6

⏭ P : P' = 1 : 4

Answer 2

Aɴꜱᴡᴇʀ

☞ In series connection power = 1.5 watt

☞ In parallel connection power = 6 watt

_________________

Gɪᴠᴇɴ

➳ Two resistors each of 12 Ω are connected in each type (series and parallel) in two different circuits.

➳ Voltage (v) = 6 V

_________________

Tᴏ ꜰɪɴᴅ

➤ Power consumed in each case

_________________

Sᴛᴇᴘꜱ

❍ First let's find the net resistance in each case,

When they are connected in series,the net resistance is given by,

$\underline { \boxed{ \blue{\sf{}R_{net }=R_1 + R_2 +R_3...... R_n}} }$

Calculating,

$\leadsto \sf{}12 \Omega + 12 \Omega \\ \\ \leadsto \pink{ \sf24 \Omega}$

When they are connected in parallel,the net resistance is given by,

$\underline{ \boxed{ \sf{} \blue{ \frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}......... \frac{1}{R_n} }}}$

Calculating,

$\leadsto \sf \frac{1}{R_{net}} = \frac{1}{12} + \frac{1}{12} \\ \\ \leadsto \sf{} \frac{1}{R_{net} } = \cancel \frac{2}{12} = \frac{1}{6} \\ \\ \leadsto \sf \pink{R_{net} = 6Ω}$

➤ So now lets calculate Power for each seperately,Power is given by,

$\underline{ \red{\boxed{{} \sf{}P = \frac{ {v}^{2} }{r}}}}$

So substituting each values,

➜ Series Connection

$\dashrightarrow \sf{}P= \frac{ {6}^{2} }{24} \\ \\ \dashrightarrow \sf{}P = \cancel \frac{36}{24} \\ \\ \orange{\dashrightarrow \sf{}P = 1.5 \: watt}$

➜ Parallel connection

$\dashrightarrow \sf{}P = \frac{ {6}^{2} }{6} \\ \\ \dashrightarrow \sf{}P = \cancel \frac{36}{6} \\ \\ \orange{ \dashrightarrow \sf{}P = 6 \: watt}$

______________________