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cricket20 [7]
1 year ago
9

An energy source will supply a constant current into the load if its internal resistance is.

Physics
1 answer:
Illusion [34]1 year ago
4 0

Answer:

If its internal resistance is zero or remains constant.

Explanation:

  • Considering the attached figure it is possible to be written an expression for the current that flows through the load connected to a source of voltage as follows.

                                                 \begin{aligned}\ \small i&=\small \frac{E}{R+r}\end{aligned}

  • Here, E,\,R,\,r represent the voltage of the energy source, the load resistance and the internal resistance of the source respectively
  • The denominator of the fraction is subject to change due to r being the possible parameter that is variable.
  • Therefore, the current can be fixed at \bold{r =0} or \bold{r = k}, k = constant.
  • In such a case the current would appear to be in either form as follows,

                                \begin{aligned}i = \frac{V}{R} \qqud \qquad \text{or}\qquad i = \frac{V}{R+k}\end{aligned}

  • Once either of the above condition is held the current  = constant.

#SPJ4

                                                 

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Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

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Explanation:

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Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

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therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
2 years ago
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