Question

question 9
in the circuit represented below, a battery of emf 30 v and unknown internal resistance r are connected to resistors, as shown. ignore the resistance of the ammeter and the connecting wires.
the current passing through the 10 ω resistor is 0.6 a.
calculate the:
3.1 equivalent resistance of the two resistors in parallel
3.2 current through the 8 ω resistor
3.3 internal resistance of the battery
(3)
(5)

Explanation:

Step1: Calculate equivalent resistance of parallel - connected resistors

The two resistors are 6Ω and 10Ω. The formula for equivalent resistance $R_{eq1}$ of two resistors $R_1$ and $R_2$ in parallel is $\frac{1}{R_{eq1}}=\frac{1}{R_1}+\frac{1}{R_2}$. Substituting $R_1 = 6\Omega$ and $R_2=10\Omega$, we have $\frac{1}{R_{eq1}}=\frac{1}{6}+\frac{1}{10}=\frac{5 + 3}{30}=\frac{8}{30}$. So, $R_{eq1}=\frac{30}{8}=3.75\Omega$.

Step2: Calculate total equivalent resistance of the circuit

The 8 - Ω resistor is in series with the parallel - combination. The total equivalent resistance $R_{total}$ of the circuit is $R_{total}=R_{eq1}+8\Omega$. Substituting $R_{eq1}=3.75\Omega$, we get $R_{total}=3.75 + 8=11.75\Omega$.

Step3: Use Ohm's law to find the current in the circuit

The voltage across the 10 - Ω resistor is $V = I\times R$, where $I = 0.6A$ and $R = 10\Omega$, so $V=0.6\times10 = 6V$. The current through the 6 - Ω resistor is $I_1=\frac{V}{6}=\frac{6}{6}=1A$. The total current in the circuit $I_{total}=I + I_1=0.6+1 = 1.6A$.

Step4: Calculate the internal resistance of the battery

The emf of the battery is $E = 30V$. Using $E=I_{total}(R_{total}+r)$, where $E = 30V$, $I_{total}=1.6A$ and $R_{total}=11.75\Omega$. We can re - arrange the formula to find $r$: $r=\frac{E}{I_{total}}-R_{total}$. Substituting the values, $r=\frac{30}{1.6}-11.75=\frac{30}{1.6}-\frac{11.75\times1.6}{1.6}=\frac{30-(11.75\times1.6)}{1.6}=\frac{30 - 18.8}{1.6}=\frac{11.2}{1.6}=7\Omega$.

Answer:

3.1: $3.75\Omega$
3.2: $1A$
3.3: $7\Omega$