Question

Two resistances 10 Ohm and 50 Ohm are connected in series with a battery of 6 volts. Calculate:

(i) The charge down from the battery per minute.
(ii) The Power dissipated in 10 Ohm resistance. ​

Answer 1

Answer:-

• R1=10ohm
• R2=50ohm

We know that in series connection

$\boxed{\sf R=R_1+R_2}$

$\\ \sf\longmapsto R=10+50$

$\\ \sf\longmapsto R=60\Omega$

• V=6v

Finding current:-

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{6}{60}$

$\\ \sf\longmapsto I=0.1A$

Finding Power

• R=10ohm
• V=6V

$\boxed{\sf P=\dfrac{V^2}{R}}$

$\\ \sf\longmapsto P=\dfrac{6^2}{10}$

$\\ \sf\longmapsto P=\dfrac{36}{10}$

$\\ \sf\longmapsto P=3.6W$

Finding Charge:-

• t=1min=60s
• I=0.1A

$\boxed{\sf I=\dfrac{Q}{t}}$

$\\ \sf\longmapsto Q=It$

$\\ \sf\longmapsto Q=0.1\times 60$

$\\ \sf\longmapsto Q=6C$

Answer 2

Answer:

R(eq)= R1+R2= 10+ 50= 60ohm

FOR CURRENT:

I=V/R= 6/60= 0.1 amp

FOR POWER AT 10 ohm resistor:

first find v at 10 ohm resistor,

V=IR=0.1×10= 1v

Now,

P=V²/R= 1x1/10=0.1 watt. ( answer a)

FOR CHARGE; (t=60 sec)

Q=IT= 0.1x60=6C. (answer b).