# Problem №7 (15 points)

Consider the power supply and measuring instruments in the presented circuit to be ideal. Determine how the readings of the instruments will change if the voltmeter and ammeter are swapped. It is known that the voltage supplied by the power supply $U_{0}=90 V$, resistances $R=50$ Ohms and $r=20$ Ohms.

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-6.jpg?height=402&width=505&top_left_y=2312&top_left_x=770)
Answer: The ammeter reading increased by $0.8 A$, and the voltmeter reading decreased by $14.3 V$

## Solution and grading criteria:

In the original circuit, the total resistance: $R_{\text {total }}=\frac{R}{2}+r=45$ Ohms.

Therefore, the total current: $I_{\text {total }}=\frac{U_{0}}{R_{\text {total }}}=2 \mathrm{~A}$.

Thus, the initial readings of the instruments: $U_{1}=I_{\text {total }} r=40$ V and $I_{1}=\frac{I_{\text {total }}}{2}=1 \mathrm{~A}$

If the instruments are swapped, then:

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-7.jpg?height=414&width=517&top_left_y=1024&top_left_x=767)

The ammeter reading: $I_{2}=\frac{U_{0}}{R}=\frac{90}{50}=1.8 \mathrm{~A}$.

The current that will flow through the resistor $r: I=\frac{U_{0}}{R+r}=\frac{90}{70}$,

Therefore, the new voltmeter reading: $U_{2}=I r=\frac{90}{70} \cdot 20 \approx 25.7$ V

Thus, the ammeter reading increased by $0.8 A$, and the voltmeter reading decreased by $14.3 V$

(3 points)