Question

in the circuit below the reading on ammeter a is 0.2 a. the battery has an emf of 9 v and internal resistance r.
5.2 calculate the current through the 5.5 ω resistor.
5.3 calculate the internal resistance of the battery

Explanation:

Step1: Analyze parallel - circuit current division

In a parallel - circuit, the voltage across each branch is the same. The two resistors in parallel are $R_1 = 11\Omega$ and $R_2=5.5\Omega$. The current through the ammeter $I = 0.2A$. Using the current - division rule for two resistors in parallel $I_2=\frac{R_1}{R_1 + R_2}I$.

Step2: Calculate current through $5.5\Omega$ resistor

Substitute $R_1 = 11\Omega$, $R_2 = 5.5\Omega$ and $I = 0.2A$ into the formula. $I_2=\frac{11}{11 + 5.5}\times0.2=\frac{11}{16.5}\times0.2=\frac{2}{16.5}\approx0.133A$.

Step3: Calculate equivalent resistance of parallel part

The equivalent resistance of the two - resistor parallel combination $R_{eq}=\frac{R_1\times R_2}{R_1 + R_2}=\frac{11\times5.5}{11 + 5.5}=\frac{60.5}{16.5}=\frac{121}{33}\Omega$.

Step4: Calculate total resistance of the circuit

The total resistance of the circuit $R_{total}=11+\frac{121}{33}+r$.

Step5: Use Ohm's law

According to Ohm's law $\epsilon=I(R_{total})$, where $\epsilon = 9V$ and $I = 0.2A$. So $9=0.2\times(11+\frac{121}{33}+r)$. First, simplify the expression inside the parentheses: $11+\frac{121}{33}=\frac{363 + 121}{33}=\frac{484}{33}\Omega$. Then the equation becomes $9 = 0.2\times(\frac{484}{33}+r)$.

Step6: Solve for internal resistance $r$

Divide both sides of the equation by $0.2$: $\frac{9}{0.2}=\frac{484}{33}+r$. $45=\frac{484}{33}+r$. Then $r = 45-\frac{484}{33}=\frac{1485-484}{33}=\frac{1001}{33}\approx30.33\Omega$.

Answer:

5.2: $0.133A$
5.3: $\frac{1001}{33}\Omega\approx30.33\Omega$