Question

In the circuit shown in the figure (in this picture), the reading on ammeter A1 is 0.4A and that on ammeter A2 is 0.64A. R1 is the internal resistance of the battery. Using the given data, calculate:
a) The resistance of R5
b) The potential difference across the terminals of the battery
c) The value of the internal resistance R1.

Answer 1

Answer:

(a) $R_5=9\ \Omega$

(b) Potential Difference = 11.584 V

(c) $R_1=0.40\ \Omega$

Explanation:

Given:

$\textrm{Current in A₁,}I_1=0.4\ A\\\textrm{Current in A₂,}I_2=0.64\ A\\R_2=5.6\ \Omega\\R_3=6.2\ \Omega\\R_4=8.2\ \Omega\\\textrm{EMF of the battery,}E= 12 V$

(a)

The resistances $R_3\ and\ R_4$ are in series. So, equivalent resistance is the sum of the two.

$R_s=R_3+R_4=8.2+6.2=14.4\ \Omega$

Now, $R_s\ and\ R_5$ are in parallel. So, potential difference across both the terminals is same. Therefore,

$I_1R_s=I_2R_5\\\\R_5=\frac{I_1}{I_2}R_s\\\\R_5=\frac{0.4}{0.64}\times 14.4=9\ \Omega$

(c)

Now, since the resistances are in parallel, the equivalent resistance is given as:

$\frac{1}{R_p}=\frac{1}{R_s}+\frac{1}{R_5}\\\\R_p=\frac{R_s\times R_5}{R_s+R_5}\\\\R_p=\frac{14.4\times 9}{14.4+9}\\\\R_p=\frac{129.6}{23.4}=5.54\ \Omega$

Now, resistances $R_1,R_2\ and\ R_p$ are in series. Therefore, equivalent resistance is given as:

$R_{eq}=R_1+R_2+R_p\\R_{eq}=R_1+5.6+5.54\\R_{eq}=R_1+11.14-----1$

Now, from Ohm's law, we know that,

$E=(I_1+I_2)R_{eq}\\\\R_{eq}=\frac{E}{I_1+I_2}\\\\R_{eq}=\frac{12}{0.4+0.64}\\\\R_{eq}=11.54\ \Omega$

Plug in $R_{eq}$ value in equation (1). This gives,

$11.54=R_1+11.14\\R_1=11.54-11.14=0.40\ \Omega$

(b)

Now, potential difference across the terminals of the battery is given as:

$V=E-(I_1+I_2)R_1\\V=12-(0.4+0.64)0.4\\V=12-0.416=11.584\ V$