Новое решение
en/8.3.22.md
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| + | ### Statement | ||
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| + | $8.3.22.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Let $r$ be the resistance of the ammeter. To double the measurement range of the meter, a shunt with resistance $r$ must be connected in parallel with the meter so that half of the measured current is diverted to the shunt. The meter and the shunt together has resistance $r/2$. Suppose that the photocell is a voltage generator $V$ and is connected to the load $R$. Without the shunt, the measured current is $V/(R+r)$. With the shunt, the measured current becomes $V/(R+r/2)$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $8.3.22.$ [Insert the problem statement] | |||
| ### Solution | |||
| Let $r$ be the resistance of the ammeter. To double the measurement range of the meter, a shunt with resistance $r$ must be connected in parallel with the meter so that half of the measured current is diverted to the shunt. The meter and the shunt together has resistance $r/2$. Suppose that the photocell is a voltage generator $V$ and is connected to the load $R$. Without the shunt, the measured current is $V/(R+r)$. With the shunt, the measured current becomes $V/(R+r/2)$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||