Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

a battery of an unknown emf and an internal resistance of 0.5 ω is conn…

Question

a battery of an unknown emf and an internal resistance of 0.5 ω is connected to three resistors, a high - resistance voltmeter and an ammeter of negligible resistance, as shown below.
the reading on the ammeter is 0.2 a.
4.1 calculate the:
4.1.1 reading on the voltmeter
4.1.2 total current supplied by the battery
4.1.3 emf of the battery
how would the voltmeter reading change if the 2 ω resistor is removed from the circuit? write down increase, decrease or remain the same.

Explanation:

Step1: Calculate voltage across 4Ω and 8Ω resistors

The 4Ω and 8Ω resistors are in series. Their equivalent resistance $R_{48}=4 + 8=12\Omega$. Using Ohm's law $V = IR$, with $I = 0.2A$, the voltage across them $V=0.2\times12 = 2.4V$. This is also the voltage across the 2Ω resistor and the volt - meter reading.

Step2: Calculate current through 2Ω resistor

Using Ohm's law $I=\frac{V}{R}$, for the 2Ω resistor with $V = 2.4V$, the current $I_{2}=\frac{2.4}{2}=1.2A$. The total current supplied by the battery $I_{total}=1.2 + 0.2=1.4A$.

Step3: Calculate emf of the battery

The total resistance in the circuit considering the internal resistance $r = 0.5\Omega$ and the equivalent resistance of the external circuit. The equivalent resistance of the external resistors: The 2Ω resistor is in parallel with the 12Ω (4Ω + 8Ω) combination. The equivalent resistance of the parallel part $R_{eq}=\frac{2\times12}{2 + 12}=\frac{24}{14}=\frac{12}{7}\Omega$. The total resistance $R_{total}=R_{eq}+r=\frac{12}{7}+0.5=\frac{12}{7}+\frac{1}{2}=\frac{24 + 7}{14}=\frac{31}{14}\Omega$. Using Ohm's law $E = I_{total}R_{total}$, with $I_{total}=1.4A$, $E=1.4\times\frac{31}{14}=3.1V$.

Step4: Analyze volt - meter reading change

If the 2Ω resistor is removed, the total resistance of the circuit increases. The total current supplied by the battery decreases. The voltage across the 4Ω and 8Ω resistors (which is the volt - meter reading) will decrease as $V = IR$ and $I$ decreases while $R$ (4Ω + 8Ω) remains the same.

Answer:

4.1.1: 2.4V
4.1.2: 1.4A
4.1.3: 3.1V
Change in volt - meter reading: DECREASE