Question

the value of p. d across 10 ohm resistance represented in an adjacent circuit is equal to​

Answer 1

Answer:

Here is the answer =>

Explanation:

As the resistors are in series, then

R effective = R1+R2+....

So R eff. = 10Ω+5Ω

= 15Ω

So Potential difference across 10Ω resistor:-

V2= (R2/R1+R2)V

= (10/5+10)*45

= 10/15*45

=2/3*45

=15*2

=30 V.

So the potential difference across 10Ω resistor is

30V

Answer 2

To find:

The value of potential difference across 10 ohm resistance represented in the given circuit.

Calculation:

5 ohm and 10 ohm resistances are in series combination such that the equivalent resistance will be:

$r_{eq} = 5 + 10 = 15 \: ohm$

So, net current through the circuit will be:

$i = \dfrac{voltage}{r_{eq}} = \dfrac{45}{15} = 3 \: amp$

So , potential drop along the 10 ohm resistor will be:

$\therefore \: V_{10\Omega} = (3 \: amp.) \times (10\Omega)$

$= > \: V_{10\Omega} = 30 \: volts$

So, final answer is:

$\boxed{ \bf{\: V_{10\Omega} = 30 \: volts}}$