Physics, 19.09.2019 14:30 laidbackkiddo412

Why do we have to connect a voltmeter in parallel and an ammeter in series

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A way that gets the desired measurement without affecting the circuit we are measuring. Voltage does not change if something is connected in parallel (assuming it doesn't draw too much current), while current doesn't change if something with low resistance is connected in series. To see this, apply Kirchoff's laws.
Kirchoff's First Law: The sum of currents flowing into a node equals the sum of currents flowing out of the node. Imagine breaking the circuit of interest at a point A and connecting an ammeter in series. Then using Kirchoff's First Law, the current flowing through the ammeter (from points A to A') is exactly the same (assuming an ideal ammeter with zero resistance) as the current that was flowing through point A before inserting the ammeter. That's precisely the behavior we want.
Kirchoff's Second Law: The sum of the voltages in any closed loop is equal to zero. Imagine connecting a voltmeter in parallel with some set of circuit components. This creates a closed loop. By Kirchoff's Second Law, the voltages in the loop must sum to zero; therefore, the voltage across the voltmeter terminals must be exactly the same as the voltage we are trying to measure. Again, this is just the behavior we are looking for.

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like poles repel so keep pushing north to north and repels funny to see this

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Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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